Let $f(x) = (x-a_1)(x-a_2)\cdots ( x-a_n), n\ge 1,$ where the $a_i$'s are distinct real numbers. For $k=0,1,\cdots, n-1$, prove that the partial fraction expansion of $\frac{x^k}{f(x)}$ is $\dfrac{x^k}{f(x)} = \sum_{i=1}^n \dfrac{a_i^k/f'(a_i)}{x-a_i}$.
I'm not sure how to prove this statement, but it seems that induction on the degree of f may be useful. When $\deg(f) = 1$, clearly equality holds. So suppose inductively that for some $n\ge 1$ and all $0\leq k < n,$ if there exist distinct real numbers $a_1,\cdots, a_n$ so that $f(x) = (x-a_1)\cdots (x-a_n)$, then $\dfrac{x^k}{f(x)} = \sum_{i=1}^n \dfrac{a_i^k/f'(a_i)}{x-a_i}$. Now consider $f(x) = (x-a_1)\cdots ( x-a_{n+1})$ for distinct real numbers $a_i$. Let $0\leq k < n + 1$. We want to show that $\frac{x^k}{f(x)} =\sum_{i=1}^{n + 1} \dfrac{a_i^k/ f'(a_i)}{x-a_i}$. Note that $\dfrac{x^k}{f(x)} = \frac{p(x)}{(x-a_1)\cdots (x-a_n)} + \frac{A}{x - a_{n+1}},$ for some polynomial $p(x)$ of degree at most $n-1$ and some constant $A$. We have $p(x)(x-a_{n+1}) + A(x-a_1)\cdots (x-a_n) = x^k$. Plugging in $x=a_{n+1}$ gives $A = \frac{a_{n+1}^k}{f'(a_{n+1})}.$ Write $p(x)= p_{n-1}x^{n-1} + \cdots + p_0.$ Then we have by induction that $\frac{p(x)}{(x-a_1)\cdots (x-a_n)} + \frac{A}{x-a_{n+1}} = \frac{a_{n+1}^k/ f'(a_{n+1})}{x-a_{n+1}} + \sum_{j=0}^{n-1} p_j \sum_{i=1}^n \frac{a_i^j/f_n'(a_i)}{x-a_i},$ where $f_n(x) = (x-a_1)\cdots ( x-a_n)$. How can I proceed from here?
Write $$\frac{x^k}{f(x)}=\sum_{i=0}^n\frac{\alpha_i}{x-a_i}$$ Then you can verify that $$\alpha_i = \lim_{x\rightarrow a_i}(x-a_i)\frac{x^k}{f(x)}=\lim_{x\rightarrow a_i}\frac{(x-a_i)}{f(x)-f(a_i)} x^k=\frac{a_i^k}{f^\prime(a_i)}$$