Proposition: Suppose that $f:[0,\infty)\to[0,\infty)$ is continuous. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $[0,\infty)$ converging to $a$. Then $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$.
Proof: Let $\epsilon>0$; since $f$ is continuous, there is a $\delta>0$ such that $|f(x)-f(x)|<\epsilon$ whenever $|x-a|<\delta$. Since $\langle x_n:n\in\Bbb N\rangle\to a$, there is an $n_0\in\Bbb N$ such that $|x_n-a|<\delta$ whenever $n\ge n_0$. But then $|f(x_n)-f(a)|<\epsilon$ whenever $n\ge n_0$, so $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$. $\dashv$
This result actually holds in much greater generality: if $X$ and $Y$ are topological spaces, $f:X\to Y$ is continuous, and $\langle x_n:n\in\Bbb N\rangle\to a$ in $X$, then $\langle f(x_n):n\in\Bbb N\rangle\to f(a)$ in $Y$; the proof is identical if $X$ and $Y$ are metric spaces and very similar even for general topological spaces. In words, continuous functions preserve convergent sequences.
Now just observe that if $p>0$, the function $f:[0,\infty)\to[0,\infty):x\mapsto x^p$ is continuous, and your generalization follows immediately.