TITLE: Find $n$, where its factorial is a product of factorials
QUESTION [26 upvotes]: 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?

REPLY [4 votes]: One can (and possibly should) be even less clever than Andreas proposes. Just compute 3! 5! 7! (getting 3628800) and then use brute force to find a number whose factorial that is. E.g., obviously the number is bigger than 7!, so start with 8! and work upwards. 8! = 40320, too small. 9! = 362880, too small. 10! = 3628800, just right -- and we're done.
If the number were much larger, requiring a search of a larger range of possible n, you could search in a marginally less-brainless way. E.g., find one n that's too small and one that's too large, then repeatedly try one in the middle and halve the range size.
Note that this requires neither intelligence nor good luck, and doesn't really have anything to do with factorials!