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.)

Obviously
$$\frac{1}{n^n}\sum_{k=1}^{n}k^k \geq 1,$$
while:
$$\frac{1}{n^n}\sum_{k=1}^{n}k^k\leq \frac{1}{n^n}\sum_{k=1}^{n}n^k\leq\frac{1}{n^n}\cdot\frac{n^n}{1-\frac{1}{n}}=\frac{n}{n-1}.$$
