Prove that if $p=2$, then taking $n$ as an even number, we have $p \mid 2^{n}-n$, and the proposition holds. Suppose $p>2$, then by Fermat's Little Theorem, we know
$$2^{p-1} \equiv 1(\bmod p)$$

Therefore, for any positive integer $k$, we have
$$2^{k(p-1)} \equiv 1(\bmod p)$$

So, it suffices to prove that there are infinitely many positive integers $k$ such that
$$\left.k(p-1) \equiv 1(\bmod p) \text { (so, let } n=k(p-1) \text {, then } p \mid 2^{n}-n\right) \text {. }$$

And this only requires $k \equiv-1(\bmod p)$, and there are certainly infinitely many such $k$.
Therefore, the proposition holds.