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?

You already have some clever answers; here's less clever approach. You want
$$
3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n,
$$
so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want 
$$
720=8\cdot9\cdots\cdot n.
$$
Now start dividing both sides by 8, then 9, etc. until you get the answer.  Dividing by 8 gives you $90=9\cdots n$, and then you either see the answer already or divide both sides by 9 to get the result.
