Proving by induction that $n! < (\frac{n+1}{2})^n$ As an analysis homework I have to prove by induction that
$n! < (\frac{n+1}{2})^n : (2 \le n \in\mathbb{N})$
For $n = 2$ this is trivial, but for $n+1$ no matter how I transform the equation I can't seem to get $(\frac{n+2}{2})^{n+1}$ on the right-hand side. I'm sure that this is an easy homework and I'm missing something blatantly obvious but the closest I got was $(n+1)! < \frac{(n+2)^{n+1}}{2^{n}}$.
Any help is greatly appreciated.
 A: By hypothesis, we have
$$\begin{align}
(n+1)!&=(n+1)n!\\\\
&<(n+1)\left(\frac{n+1}{2}\right)^n\\\\
&=2\left(\frac{n+1}{2}\right)^{n+1}\end{align}$$
From Bernoulli's Inequality, we find that 
$$\begin{align}
\left(\frac{n+2}{2}\right)^{n+1} &=\left(\frac{n+1}{2}\right)^{n+1}\left(1+\frac{1}{n+1}\right)^{n+1}\\\\
&\ge 2\left(\frac{n+1}{2}\right)^{n+1} 
\end{align}$$
And we are done!
A: $$ \frac{ \left( \frac{n+1}{2} \right)^n}{\left( \frac{n}{2} \right)^{n-1}} \; \; = \; \; \frac{1}{2} \; n \; \left(1 + \frac{1}{n} \right)^n $$
A: A full derivation. Stop here if you only are looking for a hint.
Assume $$
n! < \left(\frac{n+1}{2}\right)^n
$$
holds for some $n\geq 2$. Then,
$$
(n+1)! = (n+1)n! < (n+1)\left(\frac{n+1}{2}\right)^n
$$
by the induction hypothesis. Now, it remains to conclude to show that
$$
(n+1)\left(\frac{n+1}{2}\right)^n \leq \left(\frac{n+2}{2}\right)^{n+1}
$$
that is, rearranging, that
$$
\left(\frac{n+1}{n+2}\right)^{n+1} \leq \frac{1}{2}.
$$
But the LHS is 
$$
\left(\frac{n+1}{n+2}\right)^{n+1} = \left(1-\frac{1}{n+2}\right)^{n+1}
= e^{(n+1)\ln\left(1-\frac{1}{n+2}\right)}
\leq 
e^{-\frac{n+1}{n+2}} \leq e^{-\frac{3}{4}} < \frac{1}{2}
$$
using first that $\ln(1+x) \leq x$, and then the monotonicity of $\exp$ (and the fact that $n\geq 2$). 
A: It's worth noting that some induction proofs are more cleanly presented if you use $n-1$ for the inductive hypothesis.  That is, let's assume we have $(n-1)!\lt\left(n\over2\right)^{n-1}$.  Then
$$n!=n(n-1)!\lt n\left(n\over2\right)^{n-1}=2\left(n\over2\right)^n$$
and hence it suffices to show that
$$2\le\left(n+1\over n\right)^n=\left(1+{1\over n}\right)^n$$
This is perhaps easiest to show with a separate induction argument to establish that $(1+x)^n\ge1+nx$ for any $x$.  In this case, using $n$ for the induction hypothesis works cleanly:  If $(1+x)^n\ge1+nx$, then
$$(1+x)^{n+1}=(1+x)(1+x)^n\ge(1+x)(1+nx)=1+(n+1)x+nx^2\ge1+(n+1)x$$
Once you have $(1+x)^n\ge1+nx$ for any $x$, you can let $x={1\over n}$ and get
$$\left(1+{1\over n}\right)^n\ge1+n\cdot{1\over n}=1+1=2$$
A: Without induction
and completely elementary:
$n!
= \prod_{k=1}^n k
$
so
$\begin{array}\\
n!^2
&= (\prod_{k=1}^n k)^2\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n k)\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n (n+1-k))\\
&= \prod_{k=1}^n k(n+1-k)\\
&= \prod_{k=1}^n (k(n+1)-k^2)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\dfrac{(n+1)^2}{4}+k(n+1)-k^2\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{(n+1)^2}{4}-k(n+1)+k^2\right)\right)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\left(\dfrac{n+1}{2}-k\right)^2\right)\\
&> \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}\right)\\
&= \left(\dfrac{(n+1)^2}{4}\right)^n\\
&= \left(\dfrac{n+1}{2}\right)^{2n}\\
\text{so}\\
n!
&> \left(\dfrac{n+1}{2}\right)^{n}\\
\end{array}
$
