How to prove $S_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n$ converges to $\frac{1}{e-1}$?

Question

As the title said,I thought this sequence: $$S_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n$$ converges to $\frac{1}{e-1}$.
To prove it,I used monotonic boundedness theorem to prove it's convergent at first.
During this process,I find it's easily to use mean value inequality to obtain its monotonically decreasing property and we can used this inequality:$$(1-\frac{1}{n})^n \gt e^{-1}$$ to prove it's bounded as well.
After this,I suddenly realized there is a close connection with: $$\sum_{i=1}^{n-1} e^{-n}$$ and this series is obviously convergences to $\frac{1}{e-1}$.
In the discussion above,we already prove that:$$S_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n \lt \sum_{i=1}^{n-1} e^{-n}$$ Due to the orderliness of limits,we can get：$$\lim_{n\to\infty}S_n \leq \frac{1}{e-1}$$ Finally,we just need to get the other sides,that means to prove: $$\lim_{n\to\infty}S_n \geq \frac{1}{e-1}$$ So,how can I write a strict proof of it?