Show that $1^k+2^k+\cdots+n^k$ is $\Omega (n^{k+1})$ ...where k is a positive integer. The Big Oh case is not so hard. But how do I show that $1^k+2^k+\cdots+n^k$ is  $\Omega (n^{k+1})$?

$\frac{n^{k+1}}{k+1}=\int\limits_{0}^nx^kdx\leq1^k+2^k+\dots +n^k\leq \int\limits_{1}^{n+1}x^{k}dx=\frac{(n+1)^{k+1}-1}{k+1}$.
It is clear all monic polynomials of degree $k+1$ are asymptotically equivalent, so the result follows.
