Euler's theorem on a polynomial The following theorem is said to be due to Euler.
However, it seems not so well known.
How do you prove this?
Theorem
Let $K$ be a field.
Let $n \geq 1$ be an integer.
Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$.
Let $f(X) = (X - \alpha_1)\cdots(X - \alpha_n)$.
Then
$\sum_i \alpha_i^{k}/f'(\alpha_i) = 0$ for $0 \leq k \leq n - 2$.
$\sum_i \alpha_i^{n-1}/f'(\alpha_i) = 1$.
 A: We first work over $\mathbb{C}$.
Lemma: Let $\frac{P(z)}{Q(z)}$ be a rational function such that $Q(z)$ has distinct roots. Then the residue of $\frac{P(z)}{Q(z)}$ at $z = a$ (where $Q(a) = 0$) is $\frac{P(a)}{Q'(a)}$.
Proof. The residue is $\lim_{z \to a} \frac{P(z) (z - a)}{Q(z)} = \frac{P(a)}{Q'(a)}$ by l'Hopital's rule. $\Box$
It follows that
$$\frac{1}{f(z)} = \sum_i \frac{1}{f'(\alpha_i) (z - \alpha_i)}$$
hence
$$\frac{z}{f(z)} = \sum_i \frac{z}{f'(\alpha_i) (z - \alpha_i)} = \sum_i \frac{1}{f'(\alpha_i) \left( 1 - \frac{\alpha_i}{z} \right)}.$$
Rewriting the above identity in terms of $t = \frac{1}{z}$ gives
$$\frac{t^{n-1}}{(1 - t \alpha_1)...(1 - t \alpha_n)} = \sum_i \frac{1}{f'(\alpha_i) (1 - t \alpha_i)}$$
and comparing the first $n-1$ coefficients of the Taylor series in $t$ of both sides gives the desired result.

The final identity we prove is a sequence of polynomial identities among various symmetric polynomials in the $\alpha_i$. Consequently, if it holds over $\mathbb{C}$, it holds identically, so in particular over any field. 
A: Consider the partial fraction expansion of
$$
\frac1{f(x)}=\sum_{i=1}^n\frac{A_i}{x-\alpha_i}\tag{1}
$$
Multiply both sides by $x-\alpha_j$ for some $j$, and let $x\to\alpha_j$:
$$
\begin{align}
\lim_{x\to\alpha_j}\frac{x-\alpha_j}{f(x)}
&=\lim_{x\to\alpha_j}\sum_{i=1}^n\frac{A_i(x-\alpha_j)}{x-\alpha_i}\\
\frac1{f'(\alpha_j)}&=A_j\tag{2}
\end{align}
$$
Plugging $(2)$ into $(1)$ yields
$$
\begin{align}
\frac1{f(x)}
&=\sum_{i=1}^n\frac1{f'(\alpha_i)(x-\alpha_i)}\\
&=\sum_{i=1}^n\frac1{f'(\alpha_i)}\frac1x\left(1+\frac{\alpha_i}{x}+\frac{\alpha_i^2}{x^2}+\frac{\alpha_i^3}{x^3}+\dots\right)\tag{3}
\end{align}
$$
Multiply $(3)$ by $x^k$ and integrate around a huge counter-clockwise circle in $\mathbb{C}$.
For $k\le n-2$, the left hand side goes to $0$ since $f(x)\sim x^n$. The right hand side goes to
$$
2\pi i\sum_{i=1}^n\frac{\alpha_i^k}{f'(\alpha_i)}\tag{4}
$$
For $k=n-1$, the left hand side goes to $2\pi i$ since $f(x)\sim x^n$. The right hand side goes to
$$
2\pi i\sum_{i=1}^n\frac{\alpha_i^{n-1}}{f'(\alpha_i)}\tag{5}
$$
Therefore,
$$
\sum_{i=1}^n\frac{\alpha_i^k}{f'(\alpha_i)}=\left\{\begin{array}{}0&\mbox{if $0\le k\le n-2$}\\1&\mbox{if $k=n-1$}\end{array}\right.\tag{6}
$$
A: We consider the partial fraction decomposition of $X^{k+1}/f(X)$, where $0 \leq k \leq n - 1$.
Suppose $0 \leq k \leq n - 2$.
There exist $c_i \in K$ such that $X^{k+1}/f(X) = \sum_i c_i /(X - \alpha_i)$.
Muliplying the both sides by $f(X)$, we get
$X^{k+1} = \sum_i c_i \frac{f(X)}{X - \alpha_i}$.
Substituting $X = \alpha_i$, we get $\alpha_i^{k+1} = c_i f'(\alpha_i)$.
Hence $c_i = \alpha_i^{k+1}/f'(\alpha_i)$.
Hence $\frac{X^{k+1}}{f(X)} = \sum_i \frac{\alpha_i^{k+1}}{f'(\alpha_i)(X - \alpha_i)}$.
Substituting $X = 0$, we get
$\sum_i \frac{\alpha_i^k}{f'(\alpha_i)} = 0$.
Note that this is valid when $f(0) = 0$.
Next we consider the case $k + 1 = n$.
There exist $d_i \in K$ such that $\frac{X^n}{f(X)} = 1 + \sum_i \frac{d_i}{X - \alpha_i}$.
Muliplying the both sides by $f(X)$, we get
$X^n = f(X) + \sum_i \frac{d_i f(X)}{X - \alpha_i}$.
Substituting $X = \alpha_i$, we get $\alpha_i^n = d_i f'(\alpha_i)$.
Hence $d_i = \frac{\alpha_i^n}{f'(\alpha_i)}$.
Hence $\frac{X^n}{f(X)} = 1 + \sum_i \frac{\alpha_i^n}{f'(\alpha_i)(X - \alpha_i)}$.
Substituting $X = 0$, we get $0 = 1 - \sum_i \frac{\alpha_i^{n-1}}{f'(\alpha_i)}$.
Hence $\sum_i \frac{\alpha_i^{n-1}}{f'(\alpha_i)} = 1$.
Note that this is valid when $f(0) = 0$.
Remark
Instead of the partial fraction decompositions, we can use Lagrange's interpolation formula(e.g. the lemma 4 of my answer to this question).
A: By the the partial fraction decomposition of $\frac{1}{f(X)}$ or by Lagrange's interpolation formula(e.g. the lemma 2 of my answer to this question
), we get
$\frac{1}{f(X)} = \sum_i \frac{1}{f'(\alpha_i)(X - \alpha_i)}$
Letting $Y = \frac{1}{X}$, we get
$\frac{Y^n}{1 + b_{n-1}Y+\cdots+b_0 Y^n} = \sum_i \frac{Y}{f'(\alpha_i)(1 - \alpha_i Y)}$
,where $f(X) = X^n + b_{n-1}X^{n-1}+\cdots+b_0$.
In the ring of formal power series $K[[Y]]$,
LHS = $Y^n + c_{n+1}Y^{n+1} +\cdots$
RHS = $\sum_i \frac{Y(1 + \alpha_i Y + \alpha_i^2 Y^2 +\cdots)}{f'(\alpha_i)}$
Comparing the coefficients of $Y^k$ of the both sides, we are done.
