Is the answer of $\ \ 2415^n \ mod \ \ 2556 \ \ $ is only $711$ and $1989$ for $n\in \mathbb{Z}^{+} >1 \ \ ?$ Is the answer of  $\ \  2415^n \ mod \ \ 2556 \ \  $ is only  $711$ and $1989$ for $n\in \mathbb{Z}^{+}  >1 \ \ ?$ 
If it's true, how to prove that ?
I try to  expand  $(2556 - 141)^n $ for any $n\in \mathbb{Z}^{+}  >1$ , but it doesn't make sense.
According to WolframAlpha, 
It seems like  remainder is $711$ for $n=3,5,7,9,11,13,15,...$
and remainder is $1989$ for $n=2,4,6,8,10,12,14,16,...$
But I don't know how to prove that , or maybe the answer of  $\ \  2415^n \ mod \ \ 2556 \ \  $ is not  only  $711$ and $1989$ $?$ 
Thank you so much for every comments .
 A: We have $$1989^2\equiv 1989\ (\ mod\ 2556\ )$$ and therefore $$1989^n\equiv 1989 \ (\ mod\ 2556\ )$$ for all $n\ge 1$
That means $$2415^{2n}\equiv \ 1989\ (\ mod\ 2556\ )$$ for all $n\ge 1$
Additionally, we have $$2415\times 1989\equiv\ 711\ (\ mod\ 2556\ )$$ showing $$2415^{2n+1}\equiv 711\ (\ mod\ 2556\ )$$ for all $n\ge 1$
A: Peter has answered the question, but here is a little more to the story:
$2556$ is $4\times 639 = 2^2 \times 9\times 71 = 2^2\times 3^2 \times 71$. Therefore, by the Chinese Remainder Theorem, we can tell the behavior of powers of a number mod $2556$ by considering it mod $2^2 = 4$, mod $3^2=9$, and mod $71$.
Now $2415$ is $3=-1$ mod $4$, $3$ mod $9$, and $1$ mod $71$.
Because it is $-1$ mod $4$, its powers alternate between $-1$ and $+1$ mod $4$.
Because it is $3$ mod $9$, all its powers after the first power are $0$ mod $9$.
Because it is $1$ mod $71$, all its powers are $1$ mod $71$.
This explains the situation. The even powers are all $+1$ mod $4$, $0$ mod $9$, and $1$ mod $71$. Mod $2556$, this forces them all to be $1989$. The odd powers beyond the first are $-1$ mod $4$, $0$ mod $9$, and $1$ mod $71$. This forces them to be $711$ mod $2556$.
