Reduce $39^{13} \text{ (mod 55)}$ I am trying to reduce $39^{13} \text{ (mod 55)}$. I first tried taking successive powers of 39 to find some pattern:

$39^2\equiv36\text{ (mod 55)}$
$39^3\equiv29\text{ (mod 55)}$
$39^4\equiv31\text{ (mod 55)}$
$39^5\equiv54\text{ (mod 55)}$
$39^6\equiv16\text{ (mod 55)}$
$39^7\equiv59\text{ (mod 55)}$
$39^8\equiv46\text{ (mod 55)}$  

but it seems that it is not very helpful.

Also, 55 is composite, so we cannot make use of any theorem like Euler's Theorem.

I also tried factoring $39^{13}=3^{13}\times13^{13}$, but this way will make the calculation even longer.

Is there any better approach?

Many thanks for the help!
You missed a trick there. Note that,

$$39^5\equiv54 \equiv -1\pmod {55}$$

More generally, you can


Factor the 55 into 11 and 5 and apply CRT. This is what Euler88 did.
Factor 39 into 3 and 13 and calculate $13^{13},\    3^{13} \pmod{55}$
Note that $55=5\cdot11$, so we can use Euler's theorem mod 5 and mod 11, and then use the Chinese Remainder Theorem:
$$39^{13}\equiv39^1{\pmod 5}\equiv4{\pmod5}$$
$$39^{13}\equiv39^3{\pmod {11}}\equiv7{\pmod {11}}$$
Now the Chinese Remainder Theorem says there is a unique integer(mod 5*11) that is equivalent to $4{\pmod 5}$ and to $7{\pmod {11}}$.  In this case, that integer is $29$, so $39^{13}\equiv29{\pmod {55}}$