The last 4!

Factorial of any number $n$ where $( 20\leq n )$ has zeroes as its last four digits. So, $( 1!+2!+3!+4!+\dots+19! )$ mod $10000$ is $0313$ . So $313^{19}$ mod $10000$ is $\boxed{5977}$ .

I know it's a sin in my view but I did the problem by eliminating the options . I first found out the number mod 10 to eliminate the 3rd option . Then I just found out the last two digits. First I calculated the remainder when the entire thing is divided by 4 which is just 9^19 mod 4 $\equiv$ $1 mod 4$ . and then the entire thing $mod 25$ which is just the sum from $1!$ to $9!$ $mod 25$ (as all numbers from and after $10!$ is divisible by 25 . I wont lie, I used a calculator in this step to add the factorials from 1 to 9. which came out to be 409113 mod 25 which was obviously $\equiv$ $13 mod 25$ . and then I found out the remainder when 13^19 is divided by 25 Which is 13.19^9 $mod 25$ $\equiv$ 13.19.11^4 $mod 25$ $\equiv$ 13.19.21^2 $mod 25$ $\equiv$ -9.13.19 $mod 25$ $\equiv$ 2 $mod 25$ . So there we go the last two digits are $1 mod 4$ and $2 mod 25$ . The only option matching are when the last two digits are $77$ .