Why is true this part of Krull's height Theorem

Question

I am stuck trying to prove this part of Krull's height Theorem. I checked this proof and, in page 18, it says that if a prime ideal $\mathfrak{p}$ contains $f_1, g_2,\dots,g_n$ this implies that $f_1, f_2^{m},\dots,f_n^{m}\in\mathfrak{p}$ and hence $f_1, f_2,\dots,f_n\in\mathfrak{p}$.

It was easy to proof the first implication, but I don't know why the second one holds.

That fact is used to prove that if you quotient $R$ by $(g_2,\dots,g_n)$, then $\overline{\mathfrak{p}}$ is minimal over $(\overline{f_1})$. That's the only thing I need to (finally) understand this proof.

Answer 1

If $f_j^m$ is in a prime ideal, either $f_j^{m-1}$ or $f_j$ is in the prime ideal. Repeat this enough times to conclude that $f_j$ is in the prime ideal after all.

Answer 2

If a prime ideal $\mathfrak p$ contains some power of an element $f^m$ then it contains $f$ itself. Indeed, $f^m \in \frak p$ then $f^m = 0$ in $R/\mathfrak p$. As $\mathfrak p$ is a prime ideal, this is a domain. Hence, $f = 0$ in $R/\mathfrak p$, i.e. $f \in \mathfrak p$. An ideal satisfying this property ($f^m \in I \implies f \in I$) is called radical.