# Solving an inequality : $n \geq 3$ , $n^{n} \lt (n!)^{2}$. [duplicate]

I proved this inequality in the following way:

Lemma: $r \in \Bbb N, r \geq 3$. We have $r^r \gt (r+1)^{r-1}$.
Proof: We apply the AM-GM inequality to the $r$ positive integers where there are $r-1$ $(r+1)$'s and one $1$. We obtain $$\frac {1+(r-1)(r+1)}{r} \gt ((r+1)^{r-1})^{\frac {1}{r}}$$ wherefrom we get (since the $r^{{\rm th}}$ power function is increasing), $r^{r} \gt (r+1)^{r-1}$.

Now, I used mathematical induction to prove the statement. We have from the lemma, $(k!)^{2} \gt k^k \implies (k!)^{2} \gt (k+1)^{k-1}$ and multiplying this inequality by $(k+1)^{2}$, we have $((k+1)!)^{2} \gt (k+1)^{k+1}$ and obviously $(3!)^{2} \gt 3^3$.

Is there any direct proof of the statement that does not use induction and calculus?

## marked as duplicate by Martin Sleziak, JonMark Perry, choco_addicted, user91500, M. VinayJun 4 '16 at 8:54

Hint : $$n \le r(n-r+1)$$
Now take the product for $r=1$ to $r=n$.
• Ah... I see $(n-r)(r-1) \geq 0$. I should have seen this before. – Indrayudh Roy May 7 '14 at 20:45