But it's so large?

We can interpret "finding the last digit" as "report the value modulo 10", because the sum is finite. $\sum\limits_{i=1}^{10^{10^{10}}} i! \mod 10 = ((1! \mod 10) + (2! \mod 10) + (3! \mod 10) ...) \mod 10$ by modular arithmetic rules. We claim that for all $i \ge 5$ , $i!$ is divisible by 10, and therefore $i! \equiv 0 (\mod 10)$ for all $i \ge 5$ . We prove by induction. For the base case, we have $5! = 120$ , which is divisible by 10. For the inductive step, we assume $n!$ is divisible by 10, and then show that $(n+1)!$ 's divisibility by 10 follows. $(n+1)! = (n + 1)(n)!$ Since $n!$ is divisible by 10, we can write it as $10k$ for some $k \in \mathbb{Z}$ $(n + 1)(n)! = (n + 1)(10k)$ $(n + 1)(10k) = 10kn + 10k$ $(n + 1)(10k) = 10(kn + k)$ Since $kn + k$ is an integer, as $n, k \in \mathbb{Z}$ , we have that $(n+1)!$ is divisible by 10. Thus the induction is complete. We may now rewrite the summation as: $((1! \mod 10) + (2! \mod 10) + (3! \mod 10) + (4! \mod 10) + (0 \mod 10) + (0 \mod 10) + ....) \mod 10$ Any number of zeros added to each other still result in an overall zero answer. By simplifying the factorials, we receive $(1 + 2 + 6 + 4) \mod 10 = 13 \mod 10 = 3$