Proving identity generalizing $\frac1{(a-b)(a-c)}+\frac1{(b-a)(b-c)} + \frac1{(c-a)(c-b)} = 0$ It is simple to see $\frac{1}{a-b} + \frac{1}{b-a}=0$ and $\frac1{(a-b)(a-c)}+\frac1{(b-a)(b-c)} + \frac1{(c-a)(c-b)} = 0$, but the generalization
$$
\sum_{j=1}^n \frac{1}{(a_j-a_1) \cdots \widehat{(a_j-a_j)} \cdots (a_j-a_n)}=0$$
where the hat denotes omission, isn't as simple for me. I was unable to prove by induction. I also tried to show the derivative w.r.t. each $a_i$ vanishes, but this too is difficult. I've also tried to relate this to the determinant of a matrix with entries of the form $1/(a_i-a_j)$ but this didn't get me anywhere. 
Any help would be much appreciated. 
 A: Consider the function
$$ f(x) := \sum_{j=1}^n \frac{(x-a_1) \cdots \widehat{(x-a_j)} \cdots (x-a_n)}{(a_j-a_1) \cdots \widehat{(a_j-a_j)} \cdots (a_j-a_n)}. $$
This is a polynomial function of degree at most $n-1$.  Also, $f(a_1) = f(a_2) = \cdots = f(a_n) = 1$.  Therefore, $f(x) \equiv 1$.  If $n \ge 2$, then that implies that the coefficient of $x^{n-1}$ in this polynomial must be 0.  However, the coefficient of $x^{n-1}$ is exactly
$$\sum_{j=1}^n \frac{1}{(a_j-a_1) \cdots \widehat{(a_j-a_j)} \cdots (a_j-a_n)}.$$
A: Another proof is the following. Let
$$p(x) = (z-a_1)\cdots (z-a_n)$$
with $a_1,\ldots, a_n$ distinct. By the residue theorem we have
$$\int_{z=|R|} \frac{1}{p(z)} dz = 2\pi i \sum_{|a_j|<R} \text{Res}\left(\frac{1}{p(z)}, a_j\right).$$
If $n\geq 2$ then the absolute value of the integral is bounded by $\sim 2\pi R/R^n \to 0$ as $R\to \infty$. Since the RHS is independent of $R$ for large $R$, this implies
$$\sum_{j} \text{Res}\left(\frac{1}{p(z)}, a_j\right) = \sum_j \frac{1}{p'(a_j)}=0$$
which is equivalent to the identity. 
