TITLE: Is $77!$ divisible by $77^7$?
QUESTION [11 upvotes]: Can $77!$ be divided by $77^7$?

Attempt:
Yes, because $77=11\times 7$ and $77^7=11^7\times 7^7$ so all I need is that the prime factorization of $77!$ contains $\color{green}{11^7}\times\color{blue} {7^7}$ and it does.
$$77!=77\times...\times66\times...\times55\times...\times44\times...\times33\times...\times22\times...\times11\times...$$
and all this $\uparrow$ numbers are multiples of $11$ and there are at least $7$ so $77!$ contains for sure $\color{green}{11^7}$
And $77!$ also contains $\color{blue} {7^7}:$
$$...\times7\times...\times14\times...\times21\times...\times28\times...\times35\times...42\times...49\times...=77!$$
I have a feeling that my professor is looking for other solution.

REPLY [4 votes]: Although, the answer is already provided, I can't stress the usefulness of the Legendre's Theorem for resolving such class of problems. Especially the last one:

According to this really easy to grasp and remember result:
$$\nu_7(77!)=\frac{77-5}{7-1}=12$$
$$\nu_{11}(77!)=\frac{77-7}{11-1}=7$$
Which means $$7^{12}\cdot 11^7 \mid 77!$$