Find 3 last digits of power of power where base is huge Given $a=135797531$, find 3 last digits of $a^{a^{320}}$.
Now, I know that I need to solve: $a^{a^{320}}(\mod 1000)$.
What I know: $a\in{U_n}$ (Euler's group - Multiplicative group of integers modulo n) so that means Euler's theorem applies and $a^{\phi(1000)}\equiv1(\mod 1000)$ where $\phi(1000)=400$.
I also saw here that the exponent is congruent to the exponent modulo $\phi(1000)$ but I don't understand why and under what conditions..
I don't know how to continue from here..help!
 A: Theorem: If $b$ and $c$ are non-negative natural numbers, $m$ and $a$ are co-prime, $b = c\enspace\text{mod}\,\varphi(m)$, then $a^b = a^c\enspace\text{mod}\,m$.
Proof: Without loss of generality, suppose that $b>c$. Then $b=c+k\cdot\varphi(m)$ where $k$ is a natural number. $$a^b = a^{c+k\cdot\varphi(m)} = a^c\cdot (a^{\varphi(m)})^k = a^c \enspace\text{mod}\,m$$
Where we used Euler's theorem for $a^{\varphi(m)} = 1\enspace\text{mod}\,m$, and the fact that $$a'=a''\enspace\text{mod}\,m\,\land\,b'=b''\enspace\text{mod}\,m\implies a'b'=a''b''\enspace\text{mod}\,m$$
End of proof.
Suppose we want to calculate $a^{a^{320}}$ modulo $1000$. We can first calculate $a^{320}$ modulo $\varphi(1000)=400$ since $a$ and $1000$ are co-prime, using the theorem above. 
To calculate $a^{320}$ modulo $400$, we can calculate $320$ modulo $\varphi(400)=160$, because, again, $400$ and $a$ are co-prime.
$$320 = 0\enspace\text{mod}\,160 \implies a^{320} = 1\enspace\text{mod}\,400 \implies a^{a^{320}}=a\enspace\text{mod}\,1000 $$
And because $a=531\enspace\text{mod}\,1000$, then
$$a^{a^{320}}=531\enspace\text{mod}\,1000  $$
