What's wrong with the following solution?
Problem: Let $p$ be an odd prime number. Prove $$\binom{2p-1}{p-1} \equiv 1 \pmod{p^2}$$
Solution: It is equivalent to proving $$(2p-1)! \equiv p!(p-1)! \pmod{p^2}$$
Since $\gcd((p-1)!, p^2) = 1$, we can divide both sides by $(p-1)!$ to get $$p! \equiv p! \pmod{p^2}$$ which is obviously true.