Finding the limit of $x_n=\frac{1+2^2+\cdots+n^n}{n^n}$

Squeeze: $$ 1+2^2 + ... + n^n \leq 1+n^2 + ... + n^n = \frac{n^{n+1}-1}{n-1} $$

Whence the given sequence is dominated by $$\frac{n^{n+1} - 1}{n^n(n-1)} = \frac{n^{n+1} -1}{n^{n+1} - n^n} = 1 + \frac{n^n - 1}{n^{n+1} - n^n} = 1 + \frac{\left(1 - n^{-n}\right)}{n - 1}$$

So the limit is $1$.

Hint: try listing terms from right to left. For fixed $k \ge 0$

$$ \frac{(n-k)^{n-k}}{n^n} \le n^{-k} \ \text{if}\ n \ge k$$

Consider the sequence $z_n = \frac{1+\cdots + (n-1)^{n-1}}{n^n} = x_n -1$.

Since $x^x$ is monotonic for $x\geq 1$, we have

$$\sum_{k=1}^{n-1} k^k \leq \sum_{k=1}^{n-1} n^k < \sum_{k=0}^{n-1} k^k = \frac{n^n-1}{n-1}$$

This gives us the following string of inequalities

$$ 0 < z_n < \frac{1}{n-1}\left(1-\frac{1}{n^n}\right)$$

so $z_n \to 0$ by Squeeze theorem. Thus $x_n \to 1$.