A question about modulo with high number The question I got is 
$$
42^{27} \bmod  55
$$
And the answer I got is like this
$$
42(42^{2})^{13} \bmod  55
\\42(1764)^{13} \bmod  55
\\42(4)^{13} \bmod  55
\\42(67108864) \bmod  55
\\42(9) \bmod 55
\\378 \bmod 55
\\ 48 \bmod 55
$$
So on the third/fifth step, why $1764$ turn into $4$ and why $67108864$ turns into $9$?
I have already read others similar question but still don't get the idea about what is happening
 A: They used $\ 1764\equiv 4\pmod{55}\ $ and the Congruence Product & Power Rules.. Ditto for $\, 4^{\large 13}\!\equiv 9.\ $ 
Generally congruence is preserved by replacing aguments of sums and products by any congruent argument, e.g. see the Polynomial Congruence Rule in the above linked answer.

I just don't know how to deal with a big number 

CRT = Chinese Remainder Theorem helps by reducing to smaller moduli. Applying CRT here along with modular order reduction of exponents we obtain by easy mental arithmetic 
$\!\bmod 5\!:\ \ \ \ \ \ \ \ \ \ 2^{\large 4} \equiv 1\, \Rightarrow\ 42^{\large 27}\ \ \equiv\ \ \  2^{\large\color{#c00}{27}}\  \ \,\equiv\,\ \  2^{\large\color{#c00} 3}\ \equiv\ \color{#0a0}3,\ \ $ by $\ \ \color{#c00}{27\equiv 3}\pmod{4}$ 
$\!\bmod 11\!:\,\ (-2)^{\large 5} \equiv 1\, \Rightarrow\, 42^{\large 27 }\equiv (-2)^{\large{27}} \equiv (-2)^{\large 2} \equiv \color{#90f}4,\, \ \ $ by $\ \ 27\equiv 2\pmod{5}$ 
Thus applying easy CRT:  $ \ \ \ 42^{\large 27} \equiv  \color{#90f}4 + 11\left[\dfrac{\color{#0a0}3-\color{#90f}4}{11} \bmod 5\right]\equiv 4+11[-1]\equiv -7\equiv 48\pmod{\!55}$ 
By reducing the moduli to $5$ & $11$ we can now do the arithmetic in a couple minutes  purely mentally.
A: Another way:
Using http://mathworld.wolfram.com/CarmichaelFunction.html, 
$\lambda(55)=20$
As $(42,55)=1,27\equiv7\pmod{20}$
$42^{27}\equiv42^7\pmod{55}$
Now  $42^2\equiv4\pmod{55}$
$\implies42^7=42(42^2)^3\equiv(-13)(2^2)^3\pmod{55}$
$\equiv(-13)9\equiv-117\equiv-117+3\cdot55$
A: In the ring $\mathbb Z/55\mathbb Z$ we have
$42^{27}=(-13)^{27}=-13^{27}$
$13^2=169=4\Rightarrow13^{27}=4^{13}\cdot13=9\cdot13=7$.
Thus $$42^{27}=-7=48$$
