Proving Convergence

Question: Let $\displaystyle \{ a_n \}_{n=0}^{\infty}$ be a sequence of positive real numbers. Prove that if $\sum_{n=0}^{\infty} a_n$ converges, then so does $\sum_{n=0}^{\infty} {a_n}^r,$ where $r$ is a positive integer.

What I have tried:

I know that if  $\sum_{n=0}^{\infty} a_n$ converges, it means that $\displaystyle \{ a_n \}$ goes to 0, which is important because it tell us that when $a_n$ is less than 1, ${a_n}^{r}$ is also less than 1. 

Also, the fact that $a_n$ approaches 0 as $n$ approaches infinity allows us to conclude that there exists a positive integer $N$ for which $a_n < 1$ for all $n \ge N$, meaning that ${a_n}^r$ is also less than 1. 

Not sure what to do from here, though. :(