Finding the remainder of a factorial using modular arithmetic

Question

How do I find the remainder of $5^{2001} + (27)!$ when it is divided by $8$? Can someone please show me the appropriate steps? I'm having a hard time with modular arithmetic.

So far this is how far I've got:

$$5^2=25 \equiv 1\pmod{8} $$

So \begin{align}5^{2001}&=5^{2000}\cdot 5\\ &=(5^2)^{1000}\cdot 5\\ &=25^{1000}\cdot 5\\ &\equiv 1^{1000}\cdot 5\\ &\equiv 5 \pmod{8}\end{align}

How do I go about somthing similar for $27!$? Also, could someone direct me to a video or some notes where I can learn this?

Answer 1

$$27! = 27 \times 26\times 25 \cdots 9 \times 8\times7 \cdots 3\times2\times1=8k \; ; \quad k \in \mathbb N $$

Therefore : $$27! \equiv 0\pmod{8}$$

Also for $m \ge n$ ; $$m! \equiv 0 \pmod{n}$$