Solution:
Assume there are only finitely many such primes $p_{1}, \ldots, p_{k}$. Consider the number
$$N = 10 p_{1} p_{2} \ldots p_{k} - 1$$
Since $N$ has last digit $9$, there must be a prime $p$ dividing $N$ which does not have last digit $1$ (otherwise $N$ must have last digit $1$). But by construction, $p$ cannot divide $N$, because it leaves a remainder of $p-1$ when divided by $p$. This is a contradiction, so our assumption was wrong and there must be infinitely many such primes.