= Solution
Choose an integer $b$ whose reduction is a <primitive root> modulo $p$. At least one of $b,b+p$ satisfies $g^{p-1}\not\equiv1\pmod {p^2}$, because the <binomial theorem> gives
$$
(b+p)^{p-1}\equiv b^{p-1}+p(p-1)b^{p-2}\pmod {p^2},
$$
and the added coefficient is nonzero modulo $p$. Both choices still generate the multiplicative <group> modulo $p$. Choose such a $g$ and write $g^{p-1}=1+up$ with $p\nmid u$.
For $k\geq1$ and $p\nmid v$, another binomial expansion shows
$$
(1+vp^k)^p\equiv1+vp^{k+1}\pmod {p^{k+2}}.
$$
The intermediate binomial coefficients are divisible by $p$, and the final term also has valuation at least $k+2$ because $p$ is odd. Induction therefore gives
$$
v_p(g^{(p-1)p^j}-1)=j+1.
$$
Thus $g^{p-1}$ has order exactly $p^{n-1}$ modulo $p^n$. The order of $g$ itself is divisible by $p-1$, by reduction modulo $p$, and divides the number $p^{n-1}(p-1)$ of invertible residue classes. Its order must consequently be precisely that number. This proves the <lifting a primitive root to odd prime powers> result and
$$
\boxed{(\mathbb Z/p^n\mathbb Z)^\times\text{ is cyclic for every }n\geq1.}
$$
For the requested example choose $g=2$. Modulo $11$, $2^5\equiv-1$ and $2^2\not\equiv1$, excluding the proper divisors $1,2,5$ of ten, so its order is ten. Also $2^{10}-1=1023=11\cdot93$, which is not divisible by $11^2$. Hence
$$
\boxed{a=2\text{ generates }(\mathbb Z/11^n\mathbb Z)^\times\text{ for every }n\geq1.}
$$
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