It's easy to verify that 2 is a primitive root mod $3^2$. But then why does it follow that 2 is a primitive root mod $3^h$ for any positive integer $h$?
This was used in the solution of 2009 Putnam B6 http://math.hawaii.edu/home/pdf/putnam/Putnam_2009.pdf
I saw this Primitive roots of odd primes but unfortunately I don't have access to the book.
It's a general result that any primitive root modulo $p^2$ will serve as a primitive root for $p^k$ for any positive integer $k$. One way to show this is by lifting the required congruences using this lemma.
Suppose $g$ is a primitive root modulo $p^2$. Then it follows that $$g^{p-1} \equiv 1 \pmod{p}$$ $$g^{p-1} \not\equiv 1 \pmod{p^2}$$ Using the lemma, we can lift this into $p^3$ as $$g^{p(p-1)}\not\equiv 1 \pmod{p^3}$$
This shows that $g$ is a primitive root modulo $p^3$ since $$\mathrm{ord}(g)\mid p^2(p-1)\ \ \ \ \ \text{but}\ \ \ \ \ \mathrm{ord}(g)\nmid p(p-1)$$ and this necessarily implies that $\mathrm{ord}(g) = p^2(p-1) = \phi(p)$.
We can use the lemma to lift again into $p^4$ and then $p^5$ and so on. The same argument shown inductively will prove that $g$ is in fact a primitive modulo any $p^k$.
Lemma: $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for all $n\ge 1$.
Proof: We proceed with induction. Base case for $n=1$ is easy to verify. Suppose the lemma holds for some $n=k$. Then, by the inductive hypothesis $$2^{2\cdot 3^{k-1}}=1+3^k+3^{k+1}\ell,$$ for some $\ell\in\mathbf{Z}$. Thus, $$2^{2\cdot 3^k}=1+3^{k+1}+3^{k+2}j\equiv 1+3^{k+1}\pmod{3^{k+2}},$$ for some $j\in\mathbf{Z}$, so by induction we are done. $\Box$
Now, suppose the original proposition holds for some $n=k$, so $2^{\varphi(3^k)}=2^{2\cdot 3^{k-1}}\equiv 1\pmod{3^k}$, and let $P=\operatorname{ord}_{3^{k+1}}(2)$. Then we have $2^P\equiv 1\pmod{3^{k+1}}$, so $2^P\equiv 1\pmod{3^k}$, so $2\cdot 3^{k-1}|P$. We also know that $P|\varphi(3^{k+1})=2\cdot 3^k$, so $P=2\cdot 3^{k-1}$ or $P=2\cdot 3^k$.
Then, by our lemma, $$2^{2\cdot 3^{k-1}}\equiv 1+3^k\not\equiv 1\pmod{3^{k+1}},$$ so we must have $P=2\cdot 3^k$, and the rest follows by induction. $\Box$
Actually this problem is an application of the lifting the exponent lemma (LTE), which can be stated as follows:
Let $p$ be a prime and $n$ a nonzero integer. Then we define $v_p(n)$ to be the largest integer $k$ such that $p^k\mid n$. That is $$ v_p(n)=k\iff p^k\mid n\ \text{and} p^{k+1}\nmid n. $$
Lemma (Lift the exponent lemma). Let $p$ be a prime, $x$ and $y$ integers, and suppose that $p\mid (x-y)$ but $p\nmid x$ and $p\nmid y$. Then
(1) if $p$ is odd, $$ v_p(x^n-y^n)=v_p(x-y)+v_p(n); $$ (2) for $p=2$ and even $n$ $$ v_2(x^n-y^n)=v_2(x-y)+v_2(n)+v_2(x+y)-1. $$
Now we use the lemma to show that $2$ is a primitive root mod $3^h$ for any positive integer $h$. We need to that the smallest $n$ such $2^n=1\mod{3^h}$ is $\phi(3^h)=2\cdot 3^{h-1}$, where $\phi$ is Euler's totient function. Since $2^n=1\mod{3^h}$, we have $2^n=1\mod{3}$. This implies that $n$ is even. Write $n=2m$, so $3^h\mid (4^m-1)$. Now apply the lifting the exponent lemma: $$ h\leq v_3(4^m-1^m)=v_3(4-1)+v_3(m)=1+v_3(m). $$ So $v_3(m)\geq h-1$. The smallest possible such $m$ is $3^{h-1}$, so the smallest possible $n$ is $2\cdot 3^{h-1}=\phi(3^h)$. This proves the claim.
