A lower bound for de Polignac's formula De Polignac's Formula has many uses, for example when calculating the number of trailing zeroes of $n!$ :$$\nu_5(n!)=\sum_{i\le\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor.$$ For the purposes of my research, I'd like to find lower and upper bounds for $\nu_5(n!)$ that do not involve the floor or logarithm functions. I have found an upper one using the identity that $\lfloor\cdot\rfloor\le\cdot$ so that $$\nu_5(n!)\le n\sum_{i\le\lfloor\log_5n\rfloor}\frac1{5^i}=\frac n4\left(1-5^{\lfloor\log_5n\rfloor}\right)&lt;\frac n4$$ by geometric series. The lower one causes problems since $$\nu_5(n!)\ge\sum_{i\le\lfloor\log_5n\rfloor}\left(\frac n{5^i}-1\right)=\frac{n}{4}\left(1-5^{-\lfloor\log_5n\rfloor}\right)-\lfloor\log_5n\rfloor\ge\frac{n-1}4-\log_5n$$ gives $$\frac{n-1}4-\frac{\ln n}{\ln5}\le\nu_5(n!)&lt;\frac n4$$ and I cannot elementarily write $n$ in terms of $\nu_5(n!)$ only, for the lower bound.


  Does anyone have any further improvements on this, so that I can express $n$ as $f(\nu_5(n!))\le n&lt;g(\nu_5(n!))$ for some functions $f$ and $g$?


Edit: The approximation $\ln n\approx an^{1/a}-a$ for large $a$ is not very useful since we would essentially be handling $a$th degree polynomials.
Well one lower bound is 
$$\frac{n-4}{5}$$
 This works for all $n \ge 5$ as each time $n$ increases by $5$ an extra factor of $5$ is added to $n!$. For $n \in ${$9,14,19,24$} this lower bound is equal to the function.
You can solve $y = \frac{n-1}{4} - \frac{\ln n}{\ln 5}$ for $n$ using the Lambert W function:

$$ n = \frac{- 4 W_{-1}\left(- 5^{3/4 - y} \ln(5)/20\right)}{\ln(5)} $$