I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{} 3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\ (3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\ \end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?