Let $p$ be an odd prime and $v\geq 2$
Prove that if $g\bmod p$ has order $p-1$ in $\mathbb{Z}/p\mathbb{Z}$, then $g^{p^{v-1}}(1+p) \bmod p^v$ has order $p^v-p^{v-1}$ in $\mathbb{Z}/p^v\mathbb{Z}$
I know from another task that $1+p\mod p^v $ has order $p^{v-1}$, so $(1+p)^{p^{v-1}}\equiv 1 \pmod {p^v}$
Therefore, $$(g^{p^{v-1}}(1+p))^{p^v-p^{v-1}} =\cdots=(1+p)^{(p-1)p^{v-1}}g^{(p-1)p^{2v-2}}\equiv g^{(p-1)p^{2v-2}} \pmod{p^v}$$
Now I somehow need to show that $g^{(p-1)p^{2v-2}}\equiv 1 \pmod {p^v}$ but I'm not sure how. I can't deduce $g^{p-1}\bmod p^v$ from $g^{p-1}\bmod p$, can I?
Let $\ell=g^{p^{v-1}}(1+p)$. Note that $\varphi(p^v)=p^v-p^{v-1}$ so $\ell^{p^v-p^{v-1}}\equiv 1 \pmod{p^v}$ by Euler's theorem, so the real problem is not to show that the congruence holds but that $\varphi(p^v)$ is the order of $\ell$, i.e. $\ell$ is a primitive root modulo $p^v$.
Let $m$ be the order of $\ell$ modulo $p^v$. Now, because $\ell\equiv g^{p^{v-1}}\equiv g\pmod p$ by Fermat's theorem and since $\ell^m\equiv 1\pmod {p^{v}}$ implies $\ell^m\equiv g^m\equiv 1 \pmod p$ we have that $m$ is divisible by the order of $g$ modulo $p$, that is $m=k(p-1)$ for some non-negative integer $k$.
Now we get $$ \ell^m=\ell^{k(p-1)}=g^{k\varphi(p^v)}(p+1)^{k(p-1)}\equiv (p+1)^{k(p-1)}\equiv 1 \pmod{p^v} $$
Since the order of $p+1$ is $p^{v-1}$ modulo $p^{v}$, we get $p^{v-1}\mid k(p-1)$ and that means that $p^{v-1}(p-1)\mid m$, but we also know that $m\mid p^{v-1}(p-1)$ which implies that $m=\varphi(p^v)$, which is what we wanted to show.