Show that for any number $\displaystyle n$ and real number $\displaystyle x$ that

$\displaystyle \sum_{i=0}^n l_i(x) = 1$.

My attempt at a proof is that we should argue by induction, taking $\displaystyle n=1$ as a base case, which is easy to show:

$\displaystyle \frac{x-x_1}{x_0-x_1}+\frac{x-x_0}{x_1-x_0}$

$\displaystyle =\frac{x_1-x+x-x_0}{x_1-x_0}$

$\displaystyle =\frac{x_1-x_0}{x_1-x_0}=1$

So now let's assume that $\displaystyle \displaystyle\sum_{i=0}^n l_i(x)=1$. Then:

$\displaystyle \displaystyle\sum_{i=0}^{n+1}l_i(x)$
$\displaystyle =\displaystyle\sum_{i=0}^n l_i(x) + l_{n+1}(x)$
$\displaystyle =(x-x_{n+1})\frac{1}{????} + l_{n+1}(x)$

And it's finding out what that fraction is going to be over (as I know that sum is not just 1 due to the $\displaystyle (x_i-x_{n+1})$ terms that will be in the bottom. I think once I figure that one out the proof should fall out nicely (unless of course I'm making it way too hard and it's easier to prove directly).