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?

$$\begin{align}
3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\
&= 6 \cdot 15 \cdot 8! \\
&= 2 \cdot 5 \cdot 9! \\
&= 10 \cdot 9! \\
&= 10!
\end{align}$$
