A mosquito nuking solution relies on Taylor (Maclaurin) expansion:
$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$
and the fact that $\dfrac{\mathrm d^k}{\mathrm du^k}(u-c)^n=\begin{cases}\frac{n}{(n-k)!}(u-c)^{n-k}&k\leq n\\0&k>n\end{cases}$. Take $f(x)=p(x)$ and $x_0=0$, so that
$p^{(k)}(0)=\left.\dfrac{\mathrm d^k}{\mathrm dx^k}\sum_{i=0}^n b_i (x-c)^i\right|_{x=0}=\sum_{i=k}^n \frac{b_i i!}{(i-k)!} (-c)^{i-k}$
and use the fact that $\dbinom{i}{k}=\dfrac{i!}{k!(i-k)!}$ to obtain the answer sought. (Remember also that $\dbinom{i}{k}=0$ if i < k.)