Evaluating $\lim_{n\to\infty}\sum_{i=1}^n \left(\frac{i}{n}\right)^n$

Question

I'm experimenting with different limits and series for fun, trying to find challenging ones to evaluate, to improve my skills and knowledge. I came up with the following:

$\displaystyle\lim_{n\to\infty}\left(\sum_{i=1}^n \left(\frac{i}{n}\right)^n\right)$

AM-GM gives:

$$\sqrt[n]{1\cdot2\cdot\ldots\cdot(n-1)\cdot n}\ \leq\ \frac{1+2+\ldots+(n-1)+n}{n}$$ $$\implies 1\cdot2\cdot\ldots\cdot(n-1)\cdot n\ \leq\ \left( \frac{1}{n}+\frac{2}{n}+\ldots\frac{n-1}{n}+\frac{n}{n} \right)^n,$$

but I don't see how this is related to or can be used to help with the limit. So I then thought to do:

\begin{align}\displaystyle\sum_{i=1}^n \left(\frac{i}{n}\right)^n = 1^n + \left(1-\frac{1}{n} \right)^n+\left(1-\frac{2}{n} \right)^n + \left(1-\frac{3}{n} \right)^n + \ldots + \left(\frac{2}{n} \right)^n + \left(\frac{1}{n} \right)^n\\ \\ = 1^n + \left(1-\frac{1}{n} \right)^n + \left(\left(1-\frac{2}{n} \right)^\frac{n}{2}\right)^2 + \left(\left(1-\frac{2}{n} \right)^\frac{n}{3}\right)^3 + \ldots+\left(\frac{2}{n} \right)^n + \left(\frac{1}{n} \right)^n\\ \\ \color{red}{\approx} 1 + \frac{1}{e} + \left( \frac{1}{e} \right)^2 + \left( \frac{1}{e} \right)^3 + \ldots + \text{negligible terms} \color{red}{\approx} \frac{1}{1-\frac{1}{e}} = 1.58197\ldots.\\ \\ \end{align}

I have a feeling $\ \frac{1}{1-\frac{1}{e}}\ $ is the correct answer, but I don't feel like I've proven it because of the approximations. I had a look here, but I don't see what to use. I've got a feeling that maybe something to do with Stirling's formula could help, but I don't have much experience with it. An entirely different approach is welcome too. Thanks in advance.