The last digit

If $m\equiv{n} \pmod4$ for two positive integers $m$ and $n$ , then $2^m\equiv2^n\pmod{10}$ .

Now $4|n!$ for $n\geq4$ . Thus

$2^{1!+2!+...+100!}\equiv2^{1!+2!+3!}=2^{9}\equiv\boxed{2} \pmod{10}$

To find the last three digits, for example, we would have to analyze the exponent modulo 100 since $100=\phi(125)$ . If $m\equiv{n} \pmod{100}$ for two integers $m\geq3$ and $n\geq3$ , then $2^m\equiv2^n\pmod{1000}$ , since they are congruent both modulo 125 and modulo 8.

Now $100|n!$ for $n\geq10$ , so that

$2^{1!+2!+...+100!}\equiv2^{1!+2!+...+9!}\equiv2^{13}\equiv\boxed{192} \pmod{1000}$

If we divide $1! + 2! + 3! + \ldots + 100!$ by $4$ then we get the remainder $1$ , so, the answer is $2^1\ = 2$

Based on the observation, power of two's have period of $4$ in its last digit $2, 4, 8, 6$ . Thus, as $4 | n!$ for $n \ge 4$ , we only need to count $1! + 2! + 3! \equiv 1 \pmod{4}$ , which means the answer is $2$ .