Last non-zero digit of a factorial

The last non zero digit is equal to $5$ if and only if more $5$ 's go into $n!$ than $2$ 's. In other words $v_{5}(n!) > v_{2}(n!)$ which is impossible.

Lemma: For any $n>1$ , $2$ appears more often than $5$ in the prime factorization of $n!$ .
Proof: A prime number $p$ appears precisely $\left\lfloor\frac{n}{p}\right\rfloor + \left\lfloor\frac{n}{p^2}\right\rfloor + \cdots$ times in the prime factorization of $n!$ . Since $\left\lfloor \frac{n}{2^k} \right\rfloor \geq \left\lfloor \frac{n}{5^k} \right\rfloor$ for any $k$ , $2$ appears at least as often as $5$ as a prime factor. Furthermore, since $\left\lfloor \frac{n}{2} \right\rfloor > \left\lfloor \frac{n}{5} \right\rfloor$ (for $n>1$ ), $2$ must occur strictly more often than $5$ .

Assume that the last non-zero digit of $n!$ is $5$ . Let $n!$ have $k$ trailing zeros. Then the last digit of $\frac{n!}{10^k}$ is a $5$ , implying that $\frac{n!}{10^k} = \frac{n!}{2^k5^k}$ is odd and divisible by $5$ . Thus $2$ appears precisely $k$ times in the prime factorization of $n$ , but $5$ appears more than $k$ times, a contradiction.

Thus, $\boxed{0}$ such $n$ exist.