Road to 100!

simple! only the first four terms do matter, to get the last digit - since from 5! last digit is 0.

Therefore, last digit is the last digit of 1 + 2 + 6 + 24 i.e. 3.

Notice that for $n \geq 5$ , $n!$ has at least a trailing zero, which does not affect the unit digit of the whole sum.

Therefore, we only need to determine the unit digit of the sum of $n!$ , where $1 \leq n < 5$ to obtain the unit digit of $\displaystyle \sum_{n=1}^{100} n!$

Thus,

$\Rightarrow 1! + 2! + 3! + 4!$

$= 1 + 2 + 6 + 24 = 33$

$\Leftrightarrow \boxed 3$ is the unit digit of the above sum.