Calculate the number of 0s at the end of 2014!
Since $10=2 \times 5$, it comes down to calculating $\min \left(v_{2}(2014!), v_{5}(2014!)\right)$. However, there are more factors of 2 than factors of 5. Therefore, we calculate $v_{5}(2014!)$ using Legendre's formula:

$$
v_{5}(2014!)=\sum_{k=1}^{+\infty}\left\lfloor\frac{2014}{5^{k}}\right\rfloor=501
$$