Go ahead, use your calculator

The power $q_p(n!)$ of a prime factor $p$ in a factorial $n!$ is given by $q_p(n!) = \displaystyle \sum_{k=1}^\infty \left \lfloor \frac n {p^k} \right \rfloor$ .

Therefore, for $6!$ , we have $\begin{cases} q_2(6!) = \left \lfloor \dfrac 62 \right \rfloor + \left \lfloor \dfrac 64 \right \rfloor = 3 + 1 = 4 \\ q_3(6!) = \left \lfloor \dfrac 63 \right \rfloor = 2 \\ q_5(6!) = \left \lfloor \dfrac 65 \right \rfloor = 1 \end{cases}$

$\implies 6! = 2^4 \cdot 3^2 \cdot 5$

Similarly, for $77!$ ,

$\begin{aligned} q_2(77!) & = \left \lfloor \dfrac {77}2 \right \rfloor + \left \lfloor \dfrac {77}4 \right \rfloor + \left \lfloor \dfrac {77}8 \right \rfloor + \left \lfloor \dfrac {77}{16} \right \rfloor + \left \lfloor \dfrac {77}{32} \right \rfloor + \left \lfloor \dfrac {77}{64} \right \rfloor = 38 + 19 + 9 + 4 + 2 +1 = 73 \\ q_3(77!) & = \left \lfloor \dfrac {77}3 \right \rfloor + \left \lfloor \dfrac {77}9 \right \rfloor + \left \lfloor \dfrac {77}{27} \right \rfloor = 25+8+2=35 \\ q_5(77!) & = \left \lfloor \dfrac {77}5 \right \rfloor + \left \lfloor \dfrac {77}{25} \right \rfloor = 15+3 =18 \end{aligned}$

The maximum value of $n$ is

$\begin{aligned} n_{max} & = \min \left( \left \lfloor \dfrac {q_2(77!)}{q_2(6!)} \right \rfloor, \left \lfloor \dfrac {q_3(77!)}{q_3(6!)} \right \rfloor, \left \lfloor \dfrac {q_5(77!)}{q_5(6!)} \right \rfloor \right) \\ & = \min \left( \left \lfloor \dfrac {73}{4} \right \rfloor, \left \lfloor \dfrac {35}{2} \right \rfloor, \left \lfloor \dfrac {18}{1} \right \rfloor \right) \\ & = \min \left( 18, 17, 18 \right) \\ & = \boxed{17} \end{aligned}$