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.
 A: 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 $$7^{12}\cdot 11^7 \mid 77!$$
A: If $p$ is a prime number, the largest number $n$ such that $p^n \mid N!$ is 
$\displaystyle 
 n = \sum_{i=1}^\infty \left \lfloor \dfrac{N}{p^i}\right \rfloor$.
Note that this is really a finite series since, from some point on, all of the $\left \lfloor \dfrac{N}{p^i}\right \rfloor$ are going to be $0$. There is also a shortcut to computing $\left \lfloor \dfrac{N}{p^{i+1}}\right \rfloor$ because it can be shown that 
$$\left \lfloor \dfrac{N}{p^{i+1}}\right \rfloor =
\left \lfloor
    \dfrac{\left \lfloor \dfrac{N}
    {p^i}
\right \rfloor}{p}\right \rfloor$$
For $77!$, we get
$\qquad \left \lfloor \dfrac{77}{11}\right \rfloor = 7$
$\qquad \left \lfloor \dfrac{7}{11}\right \rfloor = 0$
So $11^7 \mid 77!$ and $11^8 \not \mid 77!$
Since $7 < 11$, it follows immediatley that $7^7 \mid 77!$. But we can also compute
$\qquad \left \lfloor \dfrac{77}{7}\right \rfloor = 11$
$\qquad \left \lfloor \dfrac{11}{7}\right \rfloor = 1$
$\qquad \left \lfloor \dfrac{1}{7}\right \rfloor = 0$
So $7^{12} \mid 77!$ and $7^{13} \not \mid 77!$
It follows that $77^7 = 7^{7} 11^7 \mid 77!$.
Added 3/9/2018
The numbers are small enough that we can show this directly
Multiples of powers of $7$ between $1$ and $77$
\begin{array}{|r|ccccccccccc|}
\hline
\text{multiple} & 7 & 14 & 21 & 28 & 35 & 42 &   49 & 56 & 63 & 70 & 77 \\
\hline
\text{power}    & 7 &  7 &  7 &  7 &  7 &  7 &  7^2 &  7 &  7 &  7 &  7 \\
\hline
\end{array}
So $7^{12} \mid 77!$.
Multiples of powers of $11$ between $1$ and $77$
\begin{array}{|r|ccccccc|}
\hline
\text{multiple} & 11 & 22 & 33 & 44 & 55 & 66 &  77\\
\hline
\text{power}    & 11 & 11 & 11 & 11 & 11 & 11 & 11  \\
\hline
\end{array}
So $11^7 \mid 77!$.
Hence $77^7 \mid 7^{12}11^7 \mid 77!$.
A: 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.
A: 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 $77!$.
