Let's say we want to determine what is the polynomial for $f(x)=(ax+b)^m$. We can get the derivatives of this:
$$
\begin{aligned}
f^{(1)}(x) & = m \cdot a \cdot (ax+b)^{m-1}\\
f^{(2)}(x) & = (m-1) \cdot m \cdot a^2 \cdot (ax+b)^{m-2}\\
f^{(n)}(x) & = (m-n+1) \cdot (m -n+20 \cdot ... \cdot (m-1) \cdot m \cdot a^n \cdot (ax+b)^{m-n}\\
& = {{m!} \over {(m-n)!}} \cdot a^n \cdot (ax+b)^{m-n}\\
\end{aligned}
$$
Getting the value of the derivatives now at zero:
$$
\begin{aligned}
f(0) & = a^0 \cdot b^m \\
f^{(1)}(0) & = {{m!} \over (m-1)!} \cdot a^1 \cdot b^{m-1}\\
f^{(2)}(0) & = {{m!} \over (m-2)!} \cdot a^2 \cdot b^{m-2}\\
f^{(n)}(0) & = {{m!} \over (m-n)!} \cdot a^n \cdot b^{m-n}\\
\end{aligned}
$$
Now we can deduce the Taylor Series:
$$
\begin{aligned}
(ax+b)^m & = \sum_{j=0}^m {{m!} \over (m-j)!} \cdot a^j \cdot b^{m-j} \cdot {{x^j} \over {j!}}\\
& = \sum_{j=0}^m {{m!} \over (m-j)!j!} \cdot a^j \cdot b^{m-j} \cdot x^j\\
& = \sum_{j=0}^m {m \choose j} \cdot a^j \cdot x^j \cdot b^{m-j}\\
\end{aligned}
$$