<s>[INST] Remainder on division with $22$ What is the remainder obtained when $14^{16}$ is divided with $22$? 
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder? 
How should I proceed?
 [/INST] Using the Euclidean Algorithm
Note that
$$
\begin{align}
14^{16}&\equiv0&\pmod2\tag1
\end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
\begin{align}
14^{16}
&\equiv3^6&\pmod{11}\\
&\equiv3&\pmod{11}\tag2
\end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&2\\\hline
1&0&1&-2\\
0&1&-5&11\\
11&2&1&0\\
\end{array}\tag3
$$
we have
$$
11(1)+2(-5)=1\tag4
$$
from which we get
$$
\begin{align}
-10&\equiv0&\pmod2\\
-10&\equiv1&\pmod{11}
\end{align}\tag5
$$
Multiplying $(5)$ by $3$ gives
$$
\begin{align}
-30&\equiv0&\pmod2\\
-30&\equiv3&\pmod{11}
\end{align}\tag6
$$

Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
\begin{align}
14&\equiv0&\pmod2\\
14&\equiv3&\pmod{11}
\end{align}\tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
\begin{align}
14^{16}&\equiv14&\pmod{22}\tag8
\end{align}
$$
</s>