= Solution
Because $(r)$ is a proper <prime ideal>, each $(r^k)$ is proper and its <radical of an ideal> is
$$
\sqrt{(r^k)}=(r)=\mathfrak p.
$$
Indeed, $(r^k)\subseteq(r)$ gives one inclusion of <radicals of ideals>, while $(rc)^k\in(r^k)$ gives the other.
We prove the saturation condition for a <primary ideal>: if $ab\in(r^k)$ and $a\notin\mathfrak p$, then $b\in(r^k)$. For $k=1$ this is the defining property of a <prime ideal>. For $k>1$, write $ab=r^kc$. Since $a\notin(r)$, the <prime ideal> condition gives $b=rb_1$. The element $r$ is a <non-zero-divisor>, so cancellation gives
$$
ab_1=r^{k-1}c.
$$
Induction gives $b_1\in(r^{k-1})$, hence $b\in(r^k)$. Together with the computed <radical of an ideal>, this proves
$$
\boxed{(r^k)\text{ is }\mathfrak p\text{-primary for every }k\geq1.}
$$
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