Showing $\left\{ \frac{1}{n^n}\sum_{k=1}^{n}{k^k} \right\}_{n \in \mathbb{N}}\rightarrow 1$ I would like to prove:
$$\left\{ \frac{1}{n^n}\sum_{k=1}^{n}{k^k} \right\}_{n \in \mathbb{N}}\rightarrow 1$$
I found a proof applying Stolz criterion but I need to use the fact that:
$$\left\{\left(\frac{n}{n+1}\right)^n\right\}\text{ is bounded}\tag{$\ast$}$$
I would like to calculate this limit without using $(\ast)$ and preferably using basic limit criterion and properties.
(Sorry about mispellings or mistakes, English is not my native language.)

Hint
$$n^n\le\sum_{k=1}^nk^k\le n^n+(n-1)^{n-1}+(n-2)(n-2)^{n-2}$$
