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?

It should work if we start factoring consecutive integers out of the expression.  You have it boiled down to this $$ 6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$
$$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$
This is a good start, but now we need an 8.  we get an 8 by using our 4 and a factor of 2 from one of our 6s.  This yields:  $$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot(3\cdot6\cdot5)$$
We can get a 9 by using our 3 and a factor of 3 from our remaining 6.  What remains:$$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot(2\cdot5)=10!$$
How about that?
