Induction proof: n! > 2^n

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Prove: $n! > 2^n$ for all $n>=4$



I already proved the base case: 24 > 16.



Then I assume this holds for n=k, and start proving it for n=k+1:



$(k+1)k! > 2^k+1$



$(k+1)k! > 2^k(2^1)$



After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.






discrete-mathematics proof-writing induction







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    You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
    – Lord Shark the Unknown
    1 hour ago






  • 2




    Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
    – user505379
    54 mins ago














up vote
3
down vote

favorite
1












Prove: $n! > 2^n$ for all $n>=4$



I already proved the base case: 24 > 16.



Then I assume this holds for n=k, and start proving it for n=k+1:



$(k+1)k! > 2^k+1$



$(k+1)k! > 2^k(2^1)$



After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.






discrete-mathematics proof-writing induction







share|cite|improve this question

















  • 2




    You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
    – Lord Shark the Unknown
    1 hour ago






  • 2




    Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
    – user505379
    54 mins ago












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Prove: $n! > 2^n$ for all $n>=4$



I already proved the base case: 24 > 16.



Then I assume this holds for n=k, and start proving it for n=k+1:



$(k+1)k! > 2^k+1$



$(k+1)k! > 2^k(2^1)$



After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.






discrete-mathematics proof-writing induction







share|cite|improve this question













Prove: $n! > 2^n$ for all $n>=4$



I already proved the base case: 24 > 16.



Then I assume this holds for n=k, and start proving it for n=k+1:



$(k+1)k! > 2^k+1$



$(k+1)k! > 2^k(2^1)$



After this I'm not sure how to proceed. Any help is appreciated, thanks in advance.






discrete-mathematics proof-writing induction




discrete-mathematics proof-writing induction






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 1 hour ago









user360431

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  • 2




    You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
    – Lord Shark the Unknown
    1 hour ago






  • 2




    Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
    – user505379
    54 mins ago












  • 2




    You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
    – Lord Shark the Unknown
    1 hour ago






  • 2




    Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
    – user505379
    54 mins ago







2




2




You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
– Lord Shark the Unknown
1 hour ago




You need to prove that $k!>2^k$ implies $(k+1)!>2^k+1$. You cannot do that by first assuming that $(k+1)!>2^k+1$.
– Lord Shark the Unknown
1 hour ago




2




2




Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
– user505379
54 mins ago




Possible duplicate of Prove the inequality $n! geq 2^n$ by induction
– user505379
54 mins ago










2 Answers
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Well, notice that



$$ (k+1)! = (k+1) k! geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2cdot2^k= 2^k+1 $$






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    The induction hypothesis is that $k!>2^k$ for some $kgeq4$.
    Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $kgeq 4$, and so $k(k+1)!>2^k+1$.






    share|cite|improve this answer




















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      2 Answers
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      up vote
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      Well, notice that



      $$ (k+1)! = (k+1) k! geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2cdot2^k= 2^k+1 $$






      share|cite|improve this answer


























        up vote
        3
        down vote













        Well, notice that



        $$ (k+1)! = (k+1) k! geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2cdot2^k= 2^k+1 $$






        share|cite|improve this answer
























          up vote
          3
          down vote










          up vote
          3
          down vote









          Well, notice that



          $$ (k+1)! = (k+1) k! geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2cdot2^k= 2^k+1 $$






          share|cite|improve this answer














          Well, notice that



          $$ (k+1)! = (k+1) k! geq (k+1) 2^k = k 2^k + 2^k > 2^k+2^k=2cdot2^k= 2^k+1 $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 58 mins ago

























          answered 1 hour ago









          Jimmy Sabater

          1,535215




          1,535215




















              up vote
              0
              down vote













              The induction hypothesis is that $k!>2^k$ for some $kgeq4$.
              Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $kgeq 4$, and so $k(k+1)!>2^k+1$.






              share|cite|improve this answer
























                up vote
                0
                down vote













                The induction hypothesis is that $k!>2^k$ for some $kgeq4$.
                Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $kgeq 4$, and so $k(k+1)!>2^k+1$.






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  The induction hypothesis is that $k!>2^k$ for some $kgeq4$.
                  Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $kgeq 4$, and so $k(k+1)!>2^k+1$.






                  share|cite|improve this answer












                  The induction hypothesis is that $k!>2^k$ for some $kgeq4$.
                  Then $(k+1)! =k!(k+1) > 2^k(k+1)$. But $2^k(k+1)>2^k2$, since $kgeq 4$, and so $k(k+1)!>2^k+1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Wuestenfux

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