Last nonzero digit of $2010!$ [closed]

$2010!$ ends with $501$ zeroes since: $$\sum_{n=1}^{5}\left\lfloor\frac{2010}{5^n}\right\rfloor=501.\tag{1}$$ For the same reason, $$ \nu_2(2010!)=\sum_{n=1}^{10}\left\lfloor\frac{2010}{2^n}\right\rfloor=2002\tag{2}$$ so $\frac{2010!}{10^{501}}$ is an even number, and since: $$ 1\cdot 2\cdot 3\cdot 4\equiv -1\pmod{5} \tag{3}$$ it follows that: $$\begin{eqnarray*}\frac{2010!}{5^{501}}\equiv \frac{(-1)^{\frac{2010}{5}}5^{402}402!}{5^{501}}&\equiv& \frac{5^{402}\cdot 2 \cdot (-1)^{\frac{400}{5}}\cdot 5^{80}\cdot 80!}{5^{501}}\equiv\frac{5^{482}\cdot 2\cdot 5^{16}\cdot 16!}{5^{501}}\\&\equiv&\frac{2\cdot 16!}{5^3}\equiv\frac{2\cdot(-1)^{\frac{15}{5}}\cdot 5^3\cdot 3!}{5^3}\equiv 3\pmod{5}\end{eqnarray*}$$ and since $2^{501}\equiv 2\pmod{5}$ we have: $$ \frac{2010!}{10^{501}}\equiv 4\pmod{5}\tag{4} $$ so the last non-zero digit of $2010!$ is $\color{red}{4}$ by the Chinese theorem.