Given prime \(p\) and a positive integer \(n\), the Legendre's formula states:
$$p^e \| n! \implies$$$$e = \left \lfloor \frac{n}{p} \right \rfloor + \left \lfloor \frac{n}{p^2} \right \rfloor + \left \lfloor \frac{n}{p^3} \right \rfloor \ldots$$
The notation [\(p^e \| n!\)] means \(p\) divides \(n!\) exactly \(e\) times (i.e. \(p^e \mid n!\) but \(p^{e+1} \nmid n!\)). This gives a lower bound:
$$e \le \frac{n}{p} + \frac{n}{p} \left ( \frac{1}{p} \right ) + \frac{n}{p} \left ( \frac{1}{p} \right )^2 + \frac{n}{p} \left ( \frac{1}{p} \right )^3 + \ldots$$
This is a geometric series, and using geometric series sum formula (click here for more info), we get:
$$e \le \frac{ \frac{n}{p} \left ( \frac{1}{p}^\infty - 1 \right ) }{\left ( \frac{1}{p} - 1 \right ) }$$
Since \( \frac{1}{p} \) is less than one:
$$e \le \frac{ \frac{n}{p} ( 0 - 1 ) }{\frac{1}{p} - 1 }$$
We can simplify this to:
$$e \le \frac{ n }{p - 1 }$$