Is $77!$ divisible by $77^7$? 
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.

Your solution is fine.  For a more general case, you could see this question for the highest power of a prime dividing a factorial.  If your professor had asked whether $2^{35}$ divides evenly into $77!$ hand counting would get very tedious.
