Prove by induction that $n! > n^2$ [duplicate]
Solution 1:
To prove that this inequality holds for $n \geq 4$, first we verify the base case, which is trivial, as $24 >16$.
Now assume for some $k$ that $k! > k^2$. Then $(k+1)! > (k+1)k^2 = k^3+k^2 > (k+1)^2$. We can verify the right hand inequality, as this implies that $k^2 > k+1 \implies k^2-k-1>0$, which is clearly true for $k \geq 4$; the inductive step has thus been proven and we're done.
Solution 2:
Hint:
If $n! > n^2$ holds, can you show that $$ (n+1)! = n! (n+1) > (n+1)^2 = n^2 +2 n+1?$$