How to determine the $\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1$. I stuck to do this,

$$\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1.$$

The only thing I have observed is $$ 1\le \lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}$$ I am unable to get its upper estimate so that I can apply Sandwich's lemma. 

$$1\le \frac{1+2^2+\ldots+n^n}{n^n}\le \frac{n+n^2+\ldots+n^n}{n^n} = \frac{n\frac{n^n-1}{n-1}}{n^n} = \frac{n^{n+1}-n}{n^{n+1}-n^n}\xrightarrow{n\to\infty} 1$$
