Say $m=17 \cdot 23 = 391$. With an exponent $e=3$ and encrypted word is $c=21$. Decrpyting exponent $d=235$. Find $w$, when $w \equiv c^{d} \pmod{m}$.
So far I have split it up like this: \begin{align*} w & \equiv 21^{235} \pmod{17}\\ w & \equiv 21^{235} \pmod{23} \end{align*}
simplified to \begin{align*} 21^{235} & \equiv 4^{235} \pmod{17}\\ 21^{235} & \equiv -2^{235} \pmod{23} \end{align*} I don't know how to further simplify and go from there. Please help!
You want \begin{align*} w& \equiv 21^{235} \equiv 4^{235} \pmod{17}\\ w& \equiv 21^{235} \equiv (-2)^{235} \equiv - (2^{235}) \pmod{23}\\ \end{align*}
Observe that $4^{2} \equiv -1 \pmod{17}$, hence $4^4 \equiv 1 \pmod{17}$. Using the division algorithm you can write $235=4(58)+3$. Thus $$4^{235} \equiv 4^{4(58)} \cdot 4^3 \equiv 1 \cdot 4^{3} \equiv -4 \equiv 13 \pmod{17}.$$
For the second congruence you may use the following $$2^{11} \equiv 1 \pmod{23}.$$ Thus $$2^{235} \equiv 2^{11(21)} \cdot 2^4 \equiv 1 \cdot 2^{4} \equiv -7 \pmod{23}.$$ Thus your system of congruences become \begin{align*} w& \equiv 13 \pmod{17}\\ w& \equiv 7 \pmod{23} \end{align*}
Now use Chinese remainder theorem to solve for $w$ (Hint: $w=30$).