Let us denote $v_p(n)$ the highest power of a prime $p$ which divides positive integer $n$. This function has obviously the following properties:
- For all primes $p$ and positive integers $n$ is $v_p(n)$ non-negative integer.
- For all positive integers $m, n$ and all primes $p$ is $v_p(mn) = v_p(m) + v_p(n)$.
- For all primes $p$ is $v_p(p!) = v_p(p) = 1$.
- For all primes $p$ is $v_p((p+1)!) = 1$, $v_p(p+1) = 0$.
- For all primes $p$ and all positive integers $n < p$ is $v_p(n!) = v_p(n) = 0$.
- Positive integer $n$ is a square if and only if $v_p(n)$ is even (including zero) for all primes $p$.

Let us denote $S = n!$ the initial value of the product on the table and $S'$ its final value after adding factorials. We can easy to see from the properties of $v_p$ that for $n$ is equal to any prime $p$ we obtain $v_p(S) = v_p(p!) = 1$ and $v_p(S') = 1$, because adding factorials does not change the amount of the prime $p$ (= $n$) in the final product on the blackboard. The number $v_p(S')$ is then odd and therefore $S'$ is not a square.

Let us assume that $n$ is a composite number (so $n \ge 4$) in whole of the following part. We will show that we can add factorials in such a way that the final product
$$S' = f_1 \cdot f_2 \cdot f_3 \cdots f_n,$$
will be a square, where $f_k$ is either $k$ or $k!$ for all $k$. It is equivalent to $v_p(S')$ is even for all primes $p$. Since $n$ is not a prime, only primes less than $n$ occur in the product $S'$. As every such primes $p$ are not in factors $f_1, f_2, \dots, f_{p-1}$ and the prime $p$ occurs in $f_p$ only once, the final power $v_p(S')$ is the same as in a “reduced” product
$$p \cdot f_{p+1} \cdot f_{p+2} \cdots f_n. \qquad (1)$$
How can we provide that every prime $p < n$ will occur in the corresponding product (1) with even power? Since in the second factor $f_{p+1}$ from (1) occurs the prime $p$ either once (in the case $f_{p+1} = (p+1)!$) or the prime $p$ does not occure (if $f_{p+1} = p+1$), we can provide “good” occurrence of $p$ by choice of $f_{p+1}$ independently on succeeding values $f_{p+2}, \dots, f_n$.

Foregoing analysis gives us construction of the required choice of factorials. Initially we choose $f_k \in \{k, k!\}$ arbitrarily for all $k \le n$ such that $k-1$ is not a prime. The other $f_k$, it is $f_{p+1}$, where $p$ is arbitrary prime less than $n$, will be chosen “backwards”, from the biggest such prime $p$ to the smallest prime $p = 2$. For the biggest unchosen $f_{p+1}$ we find parity of $v_p(f_{p+2} \dots f_n)$, in odd case we choose $f_{p+1} = p+1$, in even case we choose $f_{p+1} = (p+1)!$ and so on.

This finishes the construction of $S'$ and solution of the problem too.

Conclusion. Desired $n \ge 2$ are all composite numbers.