Prove: for any natural number n, either n is prime, n is a perfect square, or n| (n-1) !

Suppose that a natural number n is neither a prime nor a perfect square. Then, there exist natural numbers p and q such that p < n, q < n, p ≠ q and pq = n.

Now, note that (n - 1) ! = (n - 1) (n - 2) (n - 3) ... (3) (2) (1).

Since p < n and q < n, p = (n - k) and q = (n - l) where k ≠ l and k, l ∈ {∈1, 2, 3, ..., n - 1}. But we know that (n - k) (n - l) | (n - 1) !

Hence, pq | (n - 1) ! and therefore n | (n - 1) !

Hence, any natural number n is either prime, a complete square or divides (n - 1) !, as required.