real analysis - Help with filling in the details to show that $limlimits_{ntoinfty} sumlimits_{k=1}^{n}left(frac{k}{n}right)^n=frac{e}{e-1}$

So we have,

$$\begin{align} \lim_{n\to\infty} \sum_{k=1}^{n}\left(\frac{k}{n}\right)^n &= \lim_{n\to\infty} \sum_{j=0}^{n-1}\bigg(\frac{n-j}{n}\bigg)^n \\ &= \lim_{n\to\infty} \bigg(1+\bigg(1-\frac{1}{n}\bigg)^n+...+\bigg(1-\frac{n-1}{n}\bigg)^n\bigg) \\ &= 1+e^{-1}+e^{-2}+... \\ &= \frac{e}{e-1} \end{align},$$

but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $\lim\limits_{n\to\infty} \sum\limits_{k=0}^{n-1}\bigg(1+\frac{-k}{n}\bigg)^n=\sum\limits_{k=0}^\infty\lim\limits_{n\to\infty}\bigg(1+\frac{-k}{n}\bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.

Thanks in advance.