Question 1183367
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Base case:  n=4:  {{{4^2 = 16 }}} and  {{{4! = 24}}}, {{{16 <= 24 }}} so the base case holds.

Hypothesis:  Assume {{{n^2 <= n!}}} for {{{n=k }}}, {{{k>=4}}}  (*)

Step case:  Let n=k+1:

     {{{ (k+1)^2 }}}   <=(?)   {{{ (k+1)! }}}
(where (?) is provided to show that we need to resolve this inequality)

     {{{ (k+1)(k+1) }}}   <=(?)   {{{ (k+1)(k!) }}}

divide both sides by k+1:
     {{{ (k+1) }}}   <=(?)   {{{ (k!) }}}

Since {{{ (k+1) < k^2 }}}, and we have  {{{ k^2 <= k! }}} by (*), we can
write {{{ k+1 < k^2 <= k! }}} therefore  {{{ (k+1)^2 <= (k+1)! }}} and that completes the proof.