Showing that if n is a natural number larger than 3, then n ! &gt; 2 n Showing that if n is a natural number larger than 3, then n ! &gt; 2 n My try:Base Case:If n = 4, then 4 ! &gt; 2 4 24 &gt; 16So, the base case is true.Assuming P ( k ) is true. k ! &gt; 2 k Now we need to show that P ( k + 1 ) is true. ( k + 1 ) ! = 2 k + 1 Proof: ( k + 1 ) ! &gt; ( k + 1 ) k ! ⟹ ( k + 1 ) 2 k After this I have no idea how to solve further.Can anyone explain how to continue.

Question

Showing that if n is a natural number larger than 3, then   n  !  &amp;gt;      2    n  Showing that if n is a natural number larger than   3, then   n  !  &amp;gt;      2    n  My try:Base Case:If   n  =  4, then   4  !  &amp;gt;      2    4    24  &amp;gt;  16So, the base case is true.Assuming   P  (  k  ) is true.  k  !  &amp;gt;      2    k  Now we need to show that   P  (  k  +  1  ) is true.  (  k  +  1  )  !  =      2          k      +      1      Proof:  (  k  +  1  )  !  &amp;gt;  (  k  +  1  )  k  !    ⟹    (  k  +  1  )      2    k  After this I have no idea how to solve further.Can anyone explain how to continue.

Gwendolyn Case · Accepted Answer

After finding out that the base case is true and assuming   P  (  k  ) is true, for   P  (  k  +  1  ) we haveby inductive argument and since   k  +  1  &amp;gt;  2 we have   (  k  +  1  )  !  &amp;gt;      2          k      +      1