<s>[INST] Find $n$, where its factorial is a product of factorials 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?
 [/INST] We have
$$3!\cdot 5!\cdot 7!=(1\cdot 2\cdot 3)\cdot (1 \cdot 2\cdot 3\cdot 4\cdot 5)\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7,$$
and combining some of those gives
$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$ 
</s>