$g$ is a primitive root $\mod p^k$ iff it is one $\mod p^2$, thus we specify on $n=2$. Let $S$ be the set of those residues $\mod p^2$ that are primitive roots $\mod p$ and have an representant $>1$ and $<p$. Assume the contrary, meaning that there is no primitive root $g \in S$. The group of invertible residues $\mod p^2$ is cyclic (by existence of primitive roots). Especially for $d|\varphi(p^2)=p(p-1)$, there are always exactly $d$ elements $x$ with $x^d \equiv 1 \mod p^2$ (so with order dividing $d$) and exactly $\varphi(d)$ many with exact order $d$. Using $|S|=\varphi(p-1)$, we see that we already know all elements of order $(p-1)$ seen $\mod p^2$. If $s \in S$, then we look at the order $k$ of $-s \mod p^2$: $(-s)^k \equiv 1 \mod p^2 \ \iff \ s^k \equiv (-1)^k=\pm 1 \mod p^2$. But $s^k \equiv -1 \mod p^2$ happens for $k=\frac{p-1}{2}$ first (we don't need that it happens, but it does; and that it can't happen earlier follows from the fact that $s$ has order $p-1$). Especially, $k=p-1$ or $k=\frac{p-1}{2}$, the first being impossible by $-s \not\in S$ (and all with order $p-1$ are in $S$). But now we know $\varphi(p-1)$ elements of order $\frac{p-1}{2}$. This is no contradiction itself if $\varphi \left( \frac{p-1}{2} \right) = \varphi \left( p-1 \right)$ (being true iff $p \equiv 3 \mod 4$, but we will not need that). But if we just find one more, we have finally our contradiction. We know other elements of order $\frac{p-1}{2}$, namely those $s^2$ with $s \in S$. But $s \leq p-1 \implies s^2 \leq (p-1)^2=p^2-2p+1<p^2-p$, thus $s^2 \not\equiv -t \mod p^2$ for any $t \in S$. We thus found one (or in fact $|S|$ more) such elements and reached our contradiction.

ZetaX wrote: $ g$ is a primitive root $ \mod p^k$ iff it is one $ \mod p^2$, thus we specify on $ n = 2$. Why? I think 3 is primitive mod 4 but not mod 8, or do I miss something?

$ p$ is odd

For n = 1, consider a nonsquare g mod p that is not -1. $g^\frac{p-1}{2}$ = -1 mod p. Assume $g^k$ = 1 mod p with k < p-1. We know that k | p-1 however we can also see that k $\nmid \frac{p-1}{2}$ which is only possible if k = 2. But if k = 2 then g = 1, -1 which is a contradiction. Thus g is a primitive root mod p. For n > 1. Assume primitive roots exist mod $p^{n-1}$. Then these roots to the power of $p^{n-1}-p^{n-2}$ are in the set 1, $p^{n-1}+1$, $2p^{n-1}+1$,..., $(p-1)p^{n-1}+1$ mod p^n. We can see that if g is a primitive root mod $p^{n-1}$ then $(p+g)^{p^{n-1}-p^{n-2}}$ or $g^{p^{n-1}-p^{n-2}}\neq 1$ mod $p^{n}$ by the binomial theorem. So this root raised to the power of p is congruent to 1 as can be seen by the binomial theorem. If for this same root g, there existed k < $p^{n}-p^{n-1}$ such that $g^k$ = 1 mod $p^n$ then k | $p^{n-1}-p^{n-2}$ which is a contradiction.

i'll prove that there are at least $\frac{\phi(p-1)}{2}$ of such primitive roots. suppose that g is a primitive root modulo p. g' the inverse of g modulo p is a primitive root too. so if for both of them we have $g^{p-1} \equiv g'^{p-1} \equiv 1 \pmod{p^2}$ we get $p^2 | (gg')^{p-1}-1$ which is impossible because we have $p | gg'-1 , p \nmid p-1 \implies p^2 | gg'-1 < p^2-1$ so we must have gg'=1 a clear contradiction. so there are at least $\frac{\phi(p-1)}{2}$ of primitive roots which are primitive root modulo p^2 and therefor modulo p^n for all n.

This question is not related to the problem, but where did B2 go?