Inspired by Janardhanan Sivaramakrishnan

We will prove that $n!$ is a perfect number if and only if $n=3$ .

If $n=1$ , $n!=1$ is not a perfect number.

If $n=2$ , $n!=2$ is not a perfect number.

If $n=3$ , $n!=6$ is a perfect number.

If $n=4$ , $n!=24$ is not a perfect number.

If $n\geq 5$ , we have $n!$ is a even number.

We know that if $k$ is a even perfect number, $k$ always has the form $2^{p-1}(2^p-1)$ with $2^p-1$ is a prime.

Hence, $k$ has only 2 prime divisors, which are $2$ and $2^p-1$ .

If $n\geq5$ , $n!$ has at least 3 prime divisors, which are $2, 3$ and $5$ . So, $n!$ is not a perfect number.

Therefore, there is $\boxed{1}$ element of the set A is perfect number.