Is it true that $\Gamma(x+1) \leq x^x$ for every $x>0$? I've seen the inequality $\Gamma(x+1) \leq x^x$ for $x > 0$ used without proof or reference and I'm not sure it's true.
Thanks for your help.
 A: Here is a solution that only uses simple properties of the Gamma function:
For $x\geq 1$, write $x=n+t$ with $0\leq t<1$. Then by property $\Gamma(x+1)=x\Gamma(x)$, we see that
$$\Gamma(x+1)=x \Gamma((n-1)+t)=x(x-1)\Gamma((n-2)+t)=\\
x(x-1)(x-2)\cdots (x-(n-2))\Gamma(1+t)$$
It is easy to see that $\Gamma(1+t)\leq 1$ for $t\in [0,1)$, for instance by noting that for $t=0$ and $t=1$, $\Gamma(1+t)=1$ and $\Gamma(1+t)$ is convex. This shows that
$$\Gamma(x+1)\leq x(x-1)(x-2)\cdots(x-(n-2))\cdot 1\leq x^n\leq x^x$$
since there are $n$ terms in the product, each less than $x$.
A: Here is a proof relying on Euler´s integral representation of the Gamma function
$$\Gamma(z+1)=\int_{0}^{\infty}x^{z}e^{-x}dx$$
The function $x^{z}e^{\frac{-x}{2}}$ drops off to $0$ as $x\rightarrow \infty$. It´s not hard to find that its maximum is attained when $x=2z$. From this fact we can find an upper bound for $\Gamma(z+1)$ as folowing.
$$\Gamma(z+1)=\int_{0}^{\infty}x^{z}e^{-x}dx$$
$$=\int_{0}^{\infty}\left(x^{z}e^{\frac{-x}{2}}\right)e^{\frac{-x}{2}}dx$$
$$\leq\int_{0}^{\infty}(2z)^{z}e^{-z}e^{\frac{-x}{2}}dx$$
$$=(2z)^{z}e^{-z}\int_{0}^{\infty}e^{\frac{-x}{2}}dx$$
$$=2(2z)^{z}e^{-z}$$
dividing both sides of the inequality by $z^{z}$
$$\frac{\Gamma(z+1)}{z^z}\leq 2\left( \frac{2}{e}\right)^z$$
Since $2<e$, the right hand side tends to $0$ as $z\rightarrow \infty$, and we can conclude that $\frac{\Gamma(z+1)}{z^z}$.
Another more intuitive way to see it is by considering $x$ a positive integer number $n$, than
$$\Gamma(n+1)=n!= 1\cdot2\cdot3\cdots(n-1)\cdot n$$
$$n^n = \underbrace{n \cdot n \cdots n}_{n times}$$
$$\frac{n!}{n^n}=\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}$$
From the last equation it is plausible that $n^n$ grows a lot faster than $n!$!
