I believe that I saw this problem not long ago in a book:
Solve the Diophantine Equation $k!=n(n+1)(n+2)$, where $k,n$ are positive integers.
However, I am no longer able to find this question, and further examination has revealed the possibility that I may have been mistaken.
The equation appears to be similar to Brocard`s Problem, a unsolved problem in mathematics.
The only solutions appear to be $(n,k)=(1,3)(8,6)(4,5)(2,4)$.
Is there a easy way to solve this problem?
The following is slightly too long for a comment, apologies for that. Let $rad(x)$ be the largest squarefree divisor of $x$.
Conjecture (weak form of the ABC conjecture). There exists an absolute $k \in \mathbb{R}$ (probably $k = 2$ works) such that for all coprime $a, b, c$ with $a + b = c$ we have $rad(abc)^k > c$.
Theorem. Assuming the above conjecture, there are at most finitely many solutions to the equation $n! = m(m+1)(m+2)$. More precisely, for every solution we have $n < 3^{3k+1}$.
Proof. Using known bounds on the Chebyshev functions (see here), we get on the one hand,
\begin{align*} \log(rad(m(m+1)(m+2))) &= \log(rad(n!)) \\ &< 1.000028n \\ &< n\log(3) \end{align*} On the other hand, $\log(rad(m(m+2)(2m+2))) > \frac{1}{k} \log(m+1)$ by the above conjecture. We claim that these bounds contradict each other when $n > 3^{3k+1}$.
By an explicit form of Stirling's approximation, we have $\log(n!) > n (\log(n) - 1)$. Since $n! = m(m+1)(m+2) < (m+1)^3$ we get $\log(m+1) > \frac{1}{3}n (\log(n) - 1)$. We thusly obtain our desired contradiction;
\begin{align*} \log(rad(m(m+1)(m+2)) &= \log(rad(m(m+2)(2m+2))) \\ &> \frac{ \log(m+1)}{k} \\ &> \frac{n(\log(n) - 1)}{3k} \\ &> n \log(3) \end{align*}
Where the final equality uses the assumption that $n > 3^{3k+1}$.
If $k=2$ indeed works, we only have to check $3^7 = 2187$ values for $n$ to find all solutions. By being slightly more careful with estimates and using $k = 1.63$ as a value (you can find some so-called abc triples here. No example with $k \ge 1.63$ is known), only $n < 400$ have to be checked.
