how to prove $(x+1)^{(x-1)} \leq x^x$? I need to prove the statement:

$$\forall x \ge 1, \quad \log_{2}(x!) - \frac{1}{2}x\log_2(x) \geq 0$$

I tried with a proof by induction (if there is an easier way, let me know!), so I tested the basis case $x = 1$, which holds, and by induction hypothesis, assumed that $\log_{2}(x!) - \frac{1}{2}x\log_2(x) \geq 0$ holds. Now I want to come to the conclusion that $$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) \geq 0$$
After some manipulations, I find that 
$$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) = \log_{2}(x!) - \frac{1}{2}\left[x\log_2(x+1)-\log_2(x+1)\right]$$
So if $$x\log_2(x+1)-\log_2(x+1) \leq x\log_2(x)$$ then 
$$\log_{2}((x+1)!) - \frac{1}{2}(x+1)\log_2(x+1) \geq 0$$
Again, after some manipulations, I come to this inequality $(x+1)^{(x-1)} \leq x^x$. I know it is true, but I can't figure out how to prove it.
 A: Substitute $x \to x + 1$. Then, $(x+2)^{x} \leq (x+1)^{x}\cdot (x+1)\implies (1+\frac{1}{x+1})^{x} \leq x + 1$. This can be proven by induction.
The $x=1$ case is trivial, as $\frac{3}{2}\leq 2$.
Assume $(1+\frac{1}{k+1})^k \leq k+1$ is true. Then
$(1+\frac{1}{k+2})^{k+1} < (1+\frac{1}{k+1})^{k+1}\leq(k+1)(1+\frac{1}{k+1})=k+1+1=k+2$
Therefore by mathematical induction, the statement $(x+1)^{x-1}\leq x^x$ is true.
A: Your equation is equivalent to $$\left(1+\frac{1}{x}\right)^{x-1}\le x,$$ which is true for $x=1$ (since $1\le1$), for $x=2$ (since $\frac{3}{2}\le2$), and true for higher $x$ since $$\left(1+\frac{1}{x}\right)^{x-1}<e<x.$$ $\blacksquare$
A: For $x\ge 1$, examine
\begin{align}
f(x) =& x\ln x-(x-1)\ln(1+x) \\
= & f(1)+\int_1^x f’(t)dt \\
= & \>0+\int_1^x \left(f’(1)+\int_1^t f’’(s)ds\right)dt\\
  =&\int_1^x \left( 1-\ln2+\int_1^t   \frac{1-s}{s(1+s)^2}ds \right) dt\ge 0 \\
\end{align}
where $f’(1)=1-\ln2>0$ and $f’’(x)= \frac{1-x}{x(1+x)^2}\ge0 $. Thus,
$$x\ln x\ge (x-1)\ln(1+x) \implies(x+1)^{(x-1)} \leq x^x$$
A: Without induction.
Using natural logarithms, you want to prove that
$$\Delta=2 \log (x!)-x \log (x) \geq 0$$
Using Stirling approximation
$$\Delta=x (\log (x)-2)+\log (2 \pi  x)+\frac{1}{6x}+O\left(\frac{1}{x^3}\right)$$ which is positive as soon as $x \geq 2$.
