Prove or disprove: $(\frac{1}{n})^n(1 - \frac{1}{n})^{n^2-n} \simeq \frac{1}{n!}$ as $n \rightarrow \infty$.

Question

Prove or disprove: $(\frac{1}{n})^n(1 - \frac{1}{n})^{n^2-n} \simeq \frac{1}{n!}$ as $n \rightarrow \infty$.

I'm trying to prove the statement by building on my observation that $(1-\frac{1}{n})^n$ tends to $\frac{1}{e}$, which appears in Stirling's approximation for $n!$.

Answer 1

$$\frac{1}{n^n}\left(1 - \frac{1}{n}\right)^{n^2-n} \sim \frac{1}{n!} \iff \left(1 - \frac{1}{n}\right)^{n^2}\left(1 - \frac{1}{n}\right)^{-n} \sim \frac{n^n}{n!} \iff \left(1 - \frac{1}{n}\right)^{n^2} \sim \frac {e^{n-1}}{\sqrt{2\pi n}} \\ %\left(1 - \frac{1}{n}\right)^{n^2} %=\exp \left[ n^2\log \left(1-\frac 1n %\right) %\right] %=\exp \left[ n^2 \left(-\frac 1n %-\frac 1{n^2} + O\left( \frac{1}{n^3}\right) %\right)\right] %= e^{-1-n}+o(1) $$

LHS goes to $0$ and RHS to infinity.