Prove that $n! \geq n^{\frac{n}{2}}$ I was wondering how to prove that $$n! \geq n^{\frac{n}{2}}\quad\forall n \geq 1$$ 
without analytic methods that relies on asympotic comparison or use of logarithms/exponentials.
Trying by induction I was stuck immediately after the induction hypothesis because I was unable to give an estimate of the following :
$$(n+1)! = n!(n+1) \geq n^{\frac{n}{2}}(n+1)$$
Is there anything I'm missing ? Any help of tip would be appreciated, as well as other methods that don't rely on seeing the inequality as an analysis task,
I'm seeking for ''arithmetic'' or ''algebraic'' demonstrations. 
 A: For $n \ge k\ge 1$, 
note that *$$k(n-k+1)-n=(n-k)(k-1) \ge 0 \implies k(n-k+1) \ge n~~~(*)$$
Next, we write
$$n!=1 \cdot 2\cdot 3\cdot 4 \cdots k\cdots n ~~~(1)$$
and $$n!=n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots (n-k+1)\cdots 1 ~~~(2).$$
Multiplying (1) and (2) pairwise  as
$$(n!)^2=(1\cdot n)\cdot(2\cdot (n-1))\cdot (3\cdot (n-2))\cdot (4 \cdot (n-4))\cdots (k(n-k+1))\cdots (n\cdot 1)$$ 
Fo from $(*)$, it follows that 
$$(n!)^2 \ge n^n.$$
A: I think you'll need one asymptotic comparison involving a famous base of exponentials, namely $\left(1+\frac1n\right)^n\le e$. (For what I'm about to say, you only need an upper bound of $3$, which is famously easier to prove.) The ratio$$\frac{(n+1)^{(n+1)/2}}{n^{n/2}}=\left(1+\frac1n\right)^{n/2}\sqrt{n+1}\le\sqrt{e(n+1)}$$is $\le n+1$ provided $n\ge2$ (so that $n+1>e$). So you only need check the cases $n\in\{1,\,2\}$, viz. the inequalities $1!\ge1^{1/2},\,2!\ge2^{2/2}$.
A: To show that $(n+1)!\ge(n+1)^{(n+1)/2}$, it suffices to show that \begin{align}n^{n/2}(n+1)\ge(n+1)^{(n+1)/2}&\implies n^n(n+1)^2\ge(n+1)^{n+1}\\&\implies n+1\ge\left(\frac{n+1}n\right)^n\end{align} which is true for all $n\ge2$ as the RHS is never greater than $e$. To check that case $n=1$, note that $1!\ge1^{1/2}$.
A: If so, you have to prove that $$(n+1)!\geq (n+1)^{\frac{n+1}{2}}$$
We have $$n!(n+1)\geq n^{\frac{n}{2}}(n+1)\geq (n+1)^{(n+1)/2}$$ this is equivalent
$$n^{n/2}\geq (n+1)^{(n-1)/2}$$ or
$$n^n\geq (n+1)^{n-1}$$ or $$n+1\geq\left(\frac{n+1}{2}\right)^n$$ which is true.
