Lemma 2
Let $g$ be a primitive root modulo $p$ such that
$$ g^{p-1} \not \equiv 1\left(\bmod p^2\right) $$
Then for every $\alpha \geq 2$ we have
$$ g^{\varphi\left(p^{\alpha-1}\right)} \not \equiv 1\left(\bmod p^\alpha\right) . $$
Proof of Lemma 2. We use induction on $\alpha$. For $\alpha=2$, relation (9) reduces to (8). Suppose then, that (9) holds for $\alpha$. By the Euler-Fermat theorem we have
$$ g^{\varphi\left(p^{\alpha-1}\right)} \equiv 1\left(\bmod p^{\alpha-1}\right) $$
so
$$ g^{\varphi\left(p^{\alpha-1}\right)}=1+k p^{\alpha-1} $$
where $p \nmid k$ because of (9). Raising both sides of this last relation to the $p$ th power we find
$$ g^{\varphi\left(p^\alpha\right)}=\left(1+k p^{\alpha-1}\right)^p=1+k p^\alpha+k^2 \frac{p(p-1)}{2} p^{2(\alpha-1)}+r p^{3(\alpha-1)} $$
Now $2 \alpha-1 \geq \alpha+1$ and $3 \alpha-3 \geq \alpha+1$ since $\alpha \geq 2$. Hence, the last equation gives us the congruence
$$ g^{\varphi\left(p^\alpha\right)} \equiv 1+k p^\alpha\left(\bmod p^{\alpha+1}\right) $$
where $p \nmid k$. In other words, $g^{\varphi\left(p^\alpha\right)} \not \equiv 1\left(\bmod p^{\alpha+1}\right)$ so (9) holds for $\alpha+1$ if it holds for $\alpha$. This completes the proof of Lemma 2 and also of Theorem 10.6.
My question:
How can I deduce that $(g^{\phi(p^{\alpha -1})})^p=g^{\phi(p^\alpha)}$?
