I don't understand the proof of this lemma:
Let $p>1$ be an integer prime number. If $p$ is not prime gaussian integer then $p=\pi\bar{\pi}$ where $\pi$ and $\bar{\pi}$ are prime and conjugate in Gaussian integers.
Proof: Every $p \in \mathbb{Z}$ prime integer is not invertible in $Z[i]$. If $p$ is not a prime gaussian integer then there exists $\pi \in Z[i]$ prime in $Z[i]$ that divide $p$. So $\pi\bar{\pi}$ divide $p^{2}$ in $\mathbb{Z}$ so $\pi\bar{\pi} = p$. Why? I know that $\pi\bar{\pi}$ can be equal to $p^{2}$ or to $p$ or to 1. I don't know why can I say that $\pi\bar{\pi} = p$.
First, obviously, $\pi\overline{\pi} \neq 1$. Second, if $\pi\overline{\pi}=p^2$, then, taking modules, you find $|\pi|=p$, thus $p/\pi$ is a Gaussian integer of modulus one, so is a unit, and $p$ is a Gaussian prime.