elementary number theory - Fermat's little theorem and solving system of congruences

The question:

The number $561$ factors as $3 \cdot 11 \cdot 17$. First use Fermat's little theorem to prove that $$a^{561} \equiv a \pmod 3 \\ a^{561} \equiv a\pmod {11} \\ a^{561} \equiv a\pmod {17}$$ for every value of $a$. Then explain why these three congruences imply that $a^{561} \equiv a (\mod 561)$ for every value of $a$.

My attempt:

$$a^2 = \left\{ \begin{array}{c} 1 (\mod 3) \quad \text{if} \quad 3 \mid a\\ 0 (\mod 3) \quad \text{if} \quad 3 \nmid a\\ \end{array} \right. \\[3ex] a^{10} = \left\{ \begin{array}{c} 1 (\mod 11) \quad \text{if} \quad 11 \mid a\\ 0 (\mod 11) \quad \text{if} \quad 11 \nmid a\\ \end{array} \right. \\[3ex] a^{16} = \left\{ \begin{array}{c} 1 (\mod 17) \quad \text{if} \quad 17 \mid a\\ 0 (\mod 17) \quad \text{if} \quad 17 \nmid a\\ \end{array} \right.$$
I'm really not sure where to go from here. The fact that $561 = 3\cdot 11 \cdot 17$ must fit in somehow, but beyond that I don't know.