The smallest integer is
$$
k=2.
$$
Indeed, reduction modulo $p$ cannot work: the map $x\mapsto x^p$ is the identity on $\mathbb F_p^\times$, although not every element of $\mathbb Z_p^\times$ is a $p$th power.

For odd $p$, the <p-adic unit group> decomposes as
$$
\mathbb Z_p^\times\cong\mu_{p-1}\times(1+p\mathbb Z_p).
$$
Raising to the $p$th power is an automorphism on $\mu_{p-1}$. On the principal units, the <p-adic logarithm> identifies it with multiplication by $p$ on $p\mathbb Z_p$, so
$$
(1+p\mathbb Z_p)^p=1+p^2\mathbb Z_p.
$$
Hence whether a unit is a $p$th power is determined exactly by its residue modulo $p^2$. Equivalently,
$$
\alpha\in(\mathbb Z_p^\times)^p
\quad\Longleftrightarrow\quad
\alpha\bmod p^2\in((\mathbb Z/p^2\mathbb Z)^\times)^p.
$$
This is the <pth-power criterion for p-adic units>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 / 2 / c / Solution

For an odd prime $p$ and $\alpha\in\mathbb Z_p^\times$,
$$
\alpha\in(\mathbb Z_p^\times)^p
\quad\Longleftrightarrow\quad
\alpha\bmod p^2\in((\mathbb Z/p^2\mathbb Z)^\times)^p.
$$
This follows from $\mathbb Z_p^\times\cong\mu_{p-1}\times(1+p\mathbb Z_p)$ and $(1+p\mathbb Z_p)^p=1+p^2\mathbb Z_p$.


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