I think the clearest way to proceed is to use the following lifting lemma.
Lemma: Let $p$ be an odd prime and let $x$ be coprime to $p$. If for some exponent $k$ we have
$$x\equiv 1 \pmod{p^k},\ \ \ \ \ \ x \not\equiv 1 \pmod{p^{k+1}}$$
then we also have
$$x^p\equiv 1 \pmod{p^{k+1}},\ \ \ \ \ \ x^p \not\equiv 1 \pmod{p^{k+2}}$$
Proof: Let us consider the binomial theorem. Since $x\equiv 1 \pmod{p^k}$, there is some $n\in\mathbb{Z}$ such that $x = 1 + np^k$. The second congruence then requires $p\nmid n$. Taking the expression to the $p$th power then gives us
$$x^p=(1+np^k)^p = \sum_{i=0}^p\binom{p}{i}n^ip^{ki}$$
In the above expression, all terms with $i\ge 2$ are necessarily divisible by $p^{k+2}$, therefore only the first two terms survive the congruence
$$x^p \equiv 1 + np^{k+1} \not\equiv 1 \pmod{p^{k+2}}$$
but clearly, the second term is divisible by $p^{k+1}$ so we have
$$x^p \equiv 1 \pmod{p^{k+1}}$$
The result is as desired. $\square$
From your work with $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$, you know that there is some $g$ such that
$$g^{p-1} \equiv 1 \pmod p,\ \ \ \ \ \ g^{p-1} \not\equiv 1\pmod{p^2}$$
Now use the above lemma to inductively lift the congruence into $p^a$.
