Limit $\lim_{n\to \infty} \frac{1}{n^{n+1}}\sum_{k=1}^n k^p$

The reason why your approach seems to fail is because this expression is trivially just $0$. I'm assuming there's been a transcription error, but if not, then we have $$\begin{align}&\lim_{n\to\infty}\frac{1}{n^{n+1}}\sum_{k=1}^nk^p\\ =&\lim_{n\to\infty}\frac1{n^{n-p}}\left(\frac 1n\sum_{k=1}^n\left(\frac kn\right)^p\right)\end{align}$$

where the right bracket is the integral expression you're looking for (which is bounded by $1$), and the left goes to $0$.

Perhaps your sum should be $$\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^nk^p$$ in which case you just want the right bracket, which is evaluatable as a Riemann Sum.

Hint: $\int_1^{n} x^{p}dx= \sum\limits_{k=2}^{n}\int_{k-1}^{k}x^{p}dx$. Note that $\int_{k-1}^{k}x^{p}dx$ lies between $k^{p}$ and $(k-1)^{p}$. Using this, conclude that $\sum\limits_{k=1}^{n} k^{p}$ lies between $\int_1^{n} x^{p}dx+n^{p}$ and $\int_1^{n} x^{p}dx+1 -(n+1)^{p}$. Of course $\int_1^{n} x^{p}dx=\frac {n^{p+1}-1} {p+1}$. This should make it easy for you to find the limit for different values of $p$.