What is the last non-zero digit of $50!$? I'm looking for a fast method. I just multiplied all the numbers together modulo ten and divided by $5^{12}$ and $2^{12}$ modulo 10, which gave me $2$. 

Here is a fast method to determine the last non-zero digit of $n!$.

Let the base $5$ representation of $n \in\mathbb{Z}^+$ be $\overline{a_{m}a_{m-1} \ldots a_{0}}_{5}$.
Prove that the last non-zero digit of $n!$ is $\equiv 2^{\sum\limits_{i=0}^{m}ia_{i}} \times \prod\limits_{i=0}^{m}a_{i}! \pmod{10}$.

This was a question I set for some quiz in the past, and can easily be done by induction. I leave this as an exercise.
This is fast in general since $a_i!$ is small and exploiting $2^{4k+1} \equiv 2 \pmod{10}$.

Applying this to $n=50$, $50=200_5$, so the last non-zero digit of $50!$ is $$\equiv 2^{0(0)+1(0)+2(2)}(2!0!0!) \equiv 2 \pmod{10}$$

Let's apply this to $n=2013$: $2013=31023_5$ so the last non-zero digit of $2013!$ is $$\equiv 2^{0(3)+1(2)+2(0)+3(1)+4(3)}3!1!0!2!3! \equiv 2^{17}(72) \equiv 2(2) \equiv 4 \pmod{10}$$

Motivation
Well I guess I should provide some motivation. Some of this will already be in the comments. The motivation will actually be quite close to a full proof, but I've left out the more boring/easy/straightforward parts.
We may easily determine the last digit of the product of all the non-multiples of $5$ which are $\leq n$, i.e.
$$\prod_{1 \leq k \leq n, 5 \nmid k}{k}$$
by using the fact that $(5i+1)(5i+2)(5i+3)(5i+4) \equiv 4 \pmod{10}$. We get $4^j(a_0!)$ where $n=5j+a_0$. 
Intuitively, each power of $5$ sort of "removes a power of $2$" from this product, since they combine to form a power of $10$. So let's start with accounting for the multiples of $5$, and worry about multiples of $25, 125, \ldots$ later if need be.
For convenience, define $f(m)$ to be the last non-zero digit of $m$. By virtue of the fact that $n!$ is even for $n>1$, we may write at least for $n>1$
\begin{align}
f(n!) \equiv f(6^jn!) & \equiv f(6^j\prod_{1 \leq k \leq n, 5 \nmid k}{k}\prod_{1 \leq k \leq j}{5k}) \pmod{10} \\
& \equiv f\left(30^j\left(\prod_{1 \leq k \leq n, 5 \nmid k}{k}\right)j!\right) \pmod{10} \\
& \equiv f(3^j[4^j(a_0!)]j!) \pmod{10} \\
& \equiv 2^ja_0!f(j!) \pmod{10}
\end{align}
Already we can see that it is worthwile to consider the base $5$ expansion of $n$, and the term $\prod_{i=0}^{m}{a_i!}$ in the formula is apparent. What's left is to look at the product of the $2^j$ terms and express it in terms of $a_i$, and to prove the formula we get by strong induction.
