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 right on! 
You might be careful how you present it to your professor. I think it is fine, but here is other words saying the same thing. (I usually try to avoid too many $\dots$ in number theory proofs.)
$$
11 = 1\cdot 11 \\
22 = 2\cdot 11 \\
33 = 3\cdot 11 \\
44 = 4\cdot 11 \\
55 = 5\cdot 11 \\
66 = 6\cdot 11 \\
77 = 7\cdot 11
$$
are all distinct factors in $77!$, so $11^7$ divides $77!$.
Likewise 
$$
7 = 1\cdot 7 \\
14 = 2\cdot 7 \\
21 = 3\cdot 7 \\
28 = 4\cdot 7 \\
35 = 5\cdot 7 \\
42 = 6\cdot 7 \\
49 = 7\cdot 7.
$$
are factors of $77!$, so $7^7$ divide $77!$
Now since $7$ and $11$ are coprime then also are $7^7$ and $11^7$, we conclude that $7^711^7$ divide $77!$.
