# How to use Stirling's approximation in this limit?

I want to compute $$L=\lim_{n\to\infty}{\prod_{k=1}^{n}{\left(\frac{k}{n}\right)^{1/n}}}.$$ Let $$P_n$$ denote the $$n$$th partial product. Then, $$P_n=\left(\frac{n!}{n^n}\right)^{1/n}.$$ Now we can apply Stirling's approximation, $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ to obtain $$L=\frac{\sqrt{2\pi}}{e}\lim_{n\to\infty}{n^{\frac{1}{2n}}}=\frac{\sqrt{2\pi}}{e}.$$ However, the correct answer is apparently $$L=e^{-1}$$. Why am I off by a factor of $$\sqrt{2\pi}$$?

You would see the $$2\pi$$ appearing in the asymptotics but not in the limit $$P_n=\left(\frac{n!}{n^n}\right)^{1/n}=\frac{1}{e}+\frac{\log (2 \pi n)}{2 e n}+O\left(\frac{1}{n^2}\right)$$
Note that the $$\sqrt{2\pi}$$ in Stirling's approximation is just some constant, unlike the $$e$$ raised to the $$n^{\text{th}}$$ power. So, when you take $$(2\pi)^{1/n},$$ and let $$n\rightarrow\infty,$$ this $$2\pi$$ constant will go to 1 and not end up contributing to the limit.