Generalize this?

$5000! = 5000\times 4999 \times 4998 \times 4997 \times \dots \times 1$ .

Between $1$ and $5000$ , there are numbers which will be divisible by $7$ , $7^2$ , $7^3$ and $7^4$ .

There are $\large\left\lfloor\frac{500}{7}\right \rfloor = 714$ numbers that are exactly divisible by $7$ between $1$ and $5000$ .

So there are $714$ Sevens contained in these numbers.

There are $\large\left\lfloor\frac{500}{7^2}\right \rfloor = 102$ numbers that are exactly divisible by $49$ between $1$ and $5000$ .

These numbers are also a part of the $714$ numbers. But we add it again because they are divisible by $7^2$ and hence to account for the second $7$ in these numbers.

There are $\large\left\lfloor\frac{500}{7^3}\right \rfloor = 14$ numbers that are divisible by $343$ (i.e., $7^3$ ) between $1$ and $5000$ .

These numbers will be a part of the previous set $714$ and $102$ - however, we add these $14$ numbers to account for the third $7$ in these numbers as these numbers are multiples of $343$ or $7^3$ .

And finally, there are $\large\left\lfloor \frac{500}{7^4}\right \rfloor = 2$ numbers that are divisible by $2401$ (i.e. $7^4$ ).

Therefore, there will be a total of $714 + 102 + 14 + 2 = \boxed{832}$ sevens contained in $5000!$ . Hence the highest power of $7$ that can divide $5000!$ without leaving a remainder is $832$ .

In other words, the highest power of $7$ (i.e., $f(x)$ )that can divide $x!$ without leaving a remainder is:

$\begin{aligned} f(x) & =\sum _{ a=1 }^{ \infty } \left\lfloor \frac { x }{ 7^{ a } } \right\rfloor \\ f(5000) & =\sum _{ a=1 }^{ \infty } \left\lfloor \frac { 5000 }{ 7^{ a } } \right\rfloor \\ & =\left\lfloor \frac { 5000 }{ 7 } \right\rfloor +\left\lfloor \frac { 5000 }{ 7^{ 2 } } \right\rfloor +\left\lfloor \frac { 5000 }{ 7^{ 3 } } \right\rfloor +\left\lfloor \frac { 5000 }{ 7^{ 4 } } \right\rfloor \\ & =714+102+14+2 \\ & =\boxed { 832 } \end{aligned}$