Evaluation of $\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$

$$\begin{align} P &=\large\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\ &=\large\lim_{m\to\infty}\prod_{n=1}^m e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\left(\frac 12\right)^1\left(\frac 23\right)^2\left(\frac 34\right)^3\cdots \left(\frac m{m+1}\right)^m\right] \sqrt{\frac12\cdot\frac23\cdot\frac34\cdots\frac{m}{m+1}}\\ &\large=\lim_{m\to\infty}e^m\left[\frac{1\cdot 2\cdot 3\cdots m}{(m+1)^m}\color{blue}{\cdot \frac{m+1}{m+1}}\right]\sqrt\frac1{m+1}\\ &=\large\lim_{m\to\infty}e^m\frac{\color{green}{(m+1)!}}{(m+1)^{m+3/2}}\\ &\large= \lim_{m\to\infty}\frac{e^m}{(m+1)^{m+3/2}}\color{green}{\left[\sqrt{2\pi(m+1)}\left(\frac{m+1}e\right)^{m+1}\right]\quad \text{(Stirling's approx)}}\\ &\large= \frac{\sqrt{2\pi}}e\qquad \blacksquare \end{align}$$

We have: $$\log P = \sum_{n=1}^{+\infty}\left(1+\left(n+\frac{1}{2}\right)\log\left(1-\frac{1}{n+1}\right)\right)$$ but: $$\sum_{n=1}^{N}\left(1+n\log n-(n+1)\log(n+1)\right) = N-(N+1)\log(N+1)$$ since we have a telescopic sum, while: $$\sum_{n=1}^{N}\left(\frac{1}{2}\log n+\frac{1}{2}\log(n+1)\right)=\log(N!)+\frac{1}{2}\log(N+1)$$ so: $$\sum_{n=1}^{N}\left(1+\left(n+\frac{1}{2}\right)\log\left(1-\frac{1}{n+1}\right)\right)=N+\log(N!)-\left(N+\frac{1}{2}\right)\log(N+1)$$ and the result follows from the Stirling approximation: $$\log(N!) = \left(N+\frac{1}{2}\right)\log N-N+\log\sqrt{2\pi}+O\left(\frac{1}{N}\right).$$