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This was a question on my exam which I could not answer at that time (it was about 3 years ago). However, I think I have the solution now.

The prime decomposition of $2014 = 2 \cdot 19 \cdot 53$. Therefore $50! \mod 2014$ results in the following system of congruences: $$\begin{cases} 50! \equiv 0 \mod 2\\ 50! \equiv 0 \mod 19\\ 50! \equiv 50! \mod 53 \end{cases}$$ This is equivalent with $$\begin{cases} 50! \equiv 0 \mod 38\\ 50! \equiv 50! \mod 53 \end{cases}.$$

Because of Wilson's congruence, we know that $52! \equiv -1 \mod 53$. Note that $52 \cdot 51 \equiv (-1) \cdot (-2) \equiv 2 \mod 53$. Hence $2 \cdot 50! \equiv -1 \mod 53$. Mulitplying both sides with $-26$, we find that $50! \equiv 26 \mod 53$. The resulting system of congruences is $$\begin{cases} 50! \equiv 0 \mod 38\\ 50! \equiv 26 \mod 53 \end{cases}.$$

Now the inverse of $38 \mod 53$ is $7$, so we find that $26 \cdot 38 \cdot 7 = 6916$ satisfies both equations in the system of congruences. Can I conclude from the Chinese Remainder Theorem that $50! \equiv 6916 \equiv 874 \mod 2014$?

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    i have also $$50!\equiv 874 \mod 2014$$ – 2017-02-18
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    So I could make this conclusion using the Chinese Remainder Theorem? Because the system of congruences gives that $53$ divides $50! - 26$ and $38$ divides $50!$, so I would add multiples of $53$ in the first case and multiples of $38$ in the second case in order to find something like $53$ divides $n$, $38$ divides $n$ and hence $2014$ divides $n$, but this would be quite long to compute I guess... – 2017-02-18

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Yes, your proof looks correct.