4. Find the last non-zero digit of $50!=1 \times 2 \times \cdots \times 50$.
4. Notice that, in the prime factorization of 50!, 5 appears 12 times, and 2 appears more than 12 times. Therefore, the last non-zero digit of 50! is the unit digit of $\frac{50!}{2^{12} \times 5^{12}}$, and it is an even number.
Since $2^{12} \equiv 1(\bmod 5)$, we have
$$
\frac{50!}{2^{12} \times 5^{12}} \equiv \frac{50!}{5^{12}}(\bmod 5) \text {. }
$$

Thus, we only need to consider the product of the numbers in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline 1 & 2 & 3 & 4 & 1 & 6 & 7 & 8 & 9 & 2 \\
\hline 11 & 12 & 13 & 14 & 3 & 16 & 17 & 18 & 19 & 4 \\
\hline 21 & 22 & 23 & 24 & 1 & 26 & 27 & 28 & 29 & 6 \\
\hline 31 & 32 & 33 & 34 & 7 & 36 & 37 & 38 & 39 & 8 \\
\hline 41 & 42 & 43 & 44 & 9 & 46 & 47 & 48 & 49 & 2 \\
\hline
\end{tabular}

Subtracting multiples of 5 from each number, we only need to consider the product of the numbers in Table 2.
Table 2
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline 1 & 2 & 3 & 4 & 1 & 1 & 2 & 3 & 4 & 2 \\
\hline 1 & 2 & 3 & 4 & 3 & 1 & 2 & 3 & 4 & 4 \\
\hline 1 & 2 & 3 & 4 & 1 & 1 & 2 & 3 & 4 & 1 \\
\hline 1 & 2 & 3 & 4 & 2 & 1 & 2 & 3 & 4 & 3 \\
\hline 1 & 2 & 3 & 4 & 4 & 1 & 2 & 3 & 4 & 2 \\
\hline
\end{tabular}

By $4!=-1(\bmod 5)$, it is easy to see that the above product is congruent to 2 modulo 5.
Therefore, the last non-zero digit of 50! is 2.