For finding the last digit of any number we generally use,
$( n )\mod 10$ = last digit of the number.
Now for this question, $n = 1! + 2! + 3! + \ldots + 10!$, and we need to find $n^n \mod 10$.
First closely look that after $4!$ the last digit is always $0$. ( eg, $5! = 120, 6! = 720, 7! = 5040 \ldots $)
So first we add up $1! + 2! + 3! + 4! $ which is pretty trivial and straightforward and gets us $1! + 2! + 3! + 4! = 33 $ and $ 33 \mod 10 = 3$.
So this makes the question simplified as:
$$ 33^{1! + 2! + 3! + \ldots + 10!} \mod 10 $$
Proceeding,
$$ 33^{1! + 2! + 3! + \ldots + 10!} \mod 10 $$
$$ = 33^{33} \mod 10 $$
$$ = 33^{32} \cdot 33 \mod 10 $$,
$$ = 33^{8 \times 4} \cdot 33 \mod 10 = 33^{4 \times 8} \cdot 33 \mod 10 $$
$$ = 1^8 \cdot 33 \mod 10 = 33 \mod 10 $$
$$ = 3 $$
Hence the last digit is $3$.
Brute Force to verify.
I put the question directly into Wolfram, and here's the result,
$$4037913^4037913 = 1.144730641841134696286784629687671473303988068399 \ldots \times 10^{26675087} $$
and the last few decimal digits are $ \ldots 878102895\textbf{3} $
Sure do comment if you find out another way.