discrete mathematics - What is the remainder when $N = (1! + 2! + 3! + 4! + ........... + 1000! )^{40}$ is divided by $10?$

What is the remainder when $N = (1! + 2! + 3! + 4! + ........... + 1000! )^{40}$ is divided by $10$ ?

My try:

On watching the pattern as it grows, after $4!$ all are divisible by $10$.

So, infact I am just left with $N = (1! + 2! + 3! + 4! + 0)^{40}$ and I need to check the remainder when this $N$ is divisible by $10$.

Hence, the $N$ sums up to $33^{40}$ when divided by $10$ .

Now, after this I can simply apply Euler's Theorem such that

$33^{4} = 1 (mod 10)$

After all, the remainder comes out to be $1$.

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I don't have an answer for this. Is my understanding right or did I miss something?

Answer

Your answer is correct. A few pointers, however:

1. Note that you can reduce $33$ to just $3$


2. Euler's theorem says that $3^{4}\equiv 1\pmod{10}$