Prove that $\lim_{n\to\infty}\frac{(n+1)(n+2)\cdots(2n-1)2n}{n^n}=\infty$ 
Prove that $$\lim_{n\to\infty}\frac{(n+1)(n+2)\cdots(2n-1)2n}{n^n}=\infty$$

I saw that
$$\frac{(n+1)(n+2)\cdots(2n-1)2n}{n^n}=\prod_{i=1}^n\left(1+\frac in\right)$$
I think I can use squeeze law, but I'm not sure. Can you give me a hint?
 A: Consider a subset of the terms in the product's expansion. Since all such terms are positive, the original product is greater than the subset's sum:
$$\prod_{i=1}^n\left(1+\frac in\right)\ge1+\sum_{i=1}^n\frac in=1+\frac{n(n+1)}{2n}=1+\frac{n+1}2$$
Since $\frac{n+1}2$ diverges, so does the product and the original limit.
A: Hint. Note that for $n\geq 2$,
$$\frac{a_{n+1}}{a_n}=\frac{2(2n+1)}{(n+1)\left(1+\frac{1}{n}\right)^n}>\frac{2(2n+1)}{3(n+1)}\geq \frac{10}{9}.$$
A: Alternatively, using Stirling's approximation:
$$\lim_{n\to\infty}\frac{(n+1)(n+2)\cdots(2n-1)2n}{n^n}=\lim_{n\to\infty}\frac{(2n)!}{n!\cdot n^n}=\\
\lim_{n\to\infty}\frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}\cdot n^n}=\lim_{n\to\infty} \sqrt{2}\cdot \left(\frac4e\right)^n=\infty.$$
A: $e^x\geq 1+x$ holds by convexity for any $x\in\mathbb{R}$. It implies that for any $x\in(-1,1)$ we have $e^{-x}\leq \frac{1}{1+x}$ and for any $z\in(0,1)$ we have $1+z\geq e^{\frac{z}{1+z}}$. In particular
$$ \prod_{k=1}^{n-1}\left(1+\frac{k}{n}\right) \geq \exp\sum_{k=1}^{n-1}\frac{k}{n+k}=\exp\left(n-1-n\sum_{k=n+1}^{2n-1}\frac{1}{k}\right) $$
and $\sum_{k=n+1}^{2n-1}\frac{1}{k}=H_{2n-1}-H_n\leq \log(2)-\frac{2}{3n}$, hence
$$ \prod_{k=1}^{n}\left(1+\frac{k}{n}\right) \geq 2 \exp\left(n-n\log(2)-\frac{1}{3}\right)\color{red}{\geq \sqrt{2}\left(\frac{e}{2}\right)^n}. $$
A similar bound (with $\frac{e}{2}$ being replaced by $\frac{4}{e}$) can be derived by applying Stirling's double inequality
$$ \left(\frac{n}{e}\right)^n\sqrt{2\pi n}\,e^{\frac{1}{12n+1}}\leq n!\leq\left(\frac{n}{e}\right)^n\sqrt{2\pi n}\,e^{\frac{1}{12n}} $$
since
$$ \prod_{k=1}^{n}\left(1+\frac{k}{n}\right) = \frac{(2n)!}{n^n n!}.$$
