5. (10 points) For any natural number, define $n i=1 \times 2 \times 3 \times \cdots \times n$, such as $8 i=1 \times 2 \times 3 \times \cdots \times 8$; then, the expression: $2014 i+2013 i-2012 i+2011 i+\cdots-4 i+3 i-2 i+1 i$, the unit digit of the calculation result is ( )
A. 0
B. 1
C. 3
D. 9
【Solution】Solution: From the new definition $n i=1 \times 2 \times 3 \times \cdots \times n$, we know:
$$
\begin{array}{l}
2014 i=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times \cdots \times 2012 \times 2013 \times 2014 \\
2013 i=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times \cdots \times 2012 \times 2013 \\
2012 i=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times \cdots \times 2012 \\
\cdots \\
5 i=1 \times 2 \times 3 \times 4 \times 5
\end{array}
$$

By observation, it is easy to know that $2014 i, 2013 i, 2012 i, \cdots, 6 i, 5 i$ all contain the factor $2 \times 5$, so their unit digits are all 0;

Also, because:
$$
\begin{array}{l}
4 i=1 \times 2 \times 3 \times 4=24 \\
3 i=1 \times 2 \times 3=6 \\
2 i=1 \times 2=2 \\
1 i=1
\end{array}
$$

Therefore, the unit digit of $2014 i+2013 i-2012 i+2011 i+\cdots-4 i+3 i-2 i+1 i$ is: $0-4+6-2+1=1$. Hence, the answer is: B.