Show that $k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ).$ I used the following result in another post without providing proof (because I couldn't prove it):
$$k^a=\sum_{m=1}^b\left ( c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right ),$$
where $a$ and $b$ are non-negative integers, $b>a$, each $c_j\in\mathbb{C}$, each $|c_j|>0$, and each $c_j$ is unique. The product runs from $n=1$ to $b$ but skips $m$.
How can we prove this?
 A: Given $\{c_n\}_{n=1}^b$ (where $c_n \not= c_m$ for $n\not = m$) we define
$$h(z) = \prod_{n=1}^b (z-c_n)$$ Then
$$h'(c_m) = \prod_{n\not = m}^b (c_m-c_n)$$
Using this the right hand side minus the left hand side of your equality can be written
$$g(z)\equiv \sum_{m=1}^bc_m^a\frac{h(z)}{h'(c_m)(z-c_m)} - z^a$$
which is a polynomial of degree $b-1$ satisfying (since $\lim_{z\to c_m} \frac{h(z)}{z-c_m} = h'(c_m)$)
$$g(c_n) = 0,~~~~n=1,2,\ldots,b$$
thus $g(z) \equiv 0$ as no non-zero polynomial of degree $b-1$ can have more than $b-1$ roots. Note that we do not need the restriction $|c_i|>0$.
For the case $b=a+1$ there is an even simpler proof. See this related question.
A: Start by  encoding the sum  call it $S_b$  using residues. We  have by
inspection that
$$S_b = 
\sum_{m=1}^b 
\mathrm{Res}
\left(f(z); z=c_m\right)$$
where
$$f(z) = \frac{z^a}{k-z}
\prod_{n=1}^b \frac{k-c_n}{z-c_n}$$
and $c_m\ne k$ and $b>a.$

We can  therefore collect $S_b$  by integrating f(z) around  a contour
that encloses the $b+1$ poles.  We will then use  the residue at infinity
to evaluate the sum of the residues inside the contour.
Now the residue at infinity of $f(z)$ is given by
$$\mathrm{Res}
\left(-\frac{1}{z^2} f\left(\frac{1}{z}\right); z=0\right).$$
The functional term becomes
$$-\frac{1}{z^2}
\frac{1}{z^a}\frac{1}{k-1/z}
\prod_{n=1}^b \frac{k-c_n}{1/z-c_n}
= -\frac{1}{z^2}
\frac{1}{z^a}\frac{z}{zk-1} 
\prod_{n=1}^b \frac{z(k-c_n)}{1-zc_n}
\\ = -\frac{1}{z^2}
\frac{z^{b+1}}{z^a}\frac{1}{zk-1} 
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}
= \frac{1}{z^{a-b+1}}\frac{1}{1-zk} 
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$
But we have $b>a$  and hence $b-a-1 > -1$ or $b-a-1\ge  0$ so the term
is in fact
$$ z^{b-a-1} \frac{1}{1-zk} 
\prod_{n=1}^b \frac{k-c_n}{1-zc_n}.$$
and the residue at  zero of the substituted function is zero.

This means that 
$$S_b = -\mathrm{Res} \left(f(z); z=k\right)$$
which gives
$$k^a \prod_{n=1}^b \frac{k-c_n}{k-c_n} = k^a,$$
done.

Addendum I. As  an alternative to  using the residue at  infinity we
could  have used  a circular  contour  enclosing all  $b+1$ poles  and
observed that  the integral along this  contour goes to  zero since on
this circle we have
$$f(z) \in \Theta\left(R^{a-1} \times R^{-b}\right)
= \Theta\left(R^{a-b-1}\right)$$
and
$$\lim_{R\to\infty} 2\pi R \times R^{a-b-1}
= 2\pi \lim_{R\to\infty} R^{a-b} = 0$$
because $b>a.$
This has the advantage of working for $a$ a positive real parameter as opposed to a positive integer only.
Addendum II. The case $k=c_m$ for some $m$ gives
$$\sum_{m=1}^b\left(c_m^a\prod_{n\neq m} \frac{k-c_n}{c_m-c_n} \right)
= k^a$$ because all the products vanish except the one for $c_m = k.$

A similar calculation can be found at this MSE link.
