How to show this Lagrange Polynomial formula?

Question

For a Lagrange Polynomial, how to show

$$l_0(x) = 1 + \frac{x-x_0}{x_0 - x_1} + \frac{(x-x_0)(x-x_1)}{(x_0 - x_1)(x_0 - x_2)} + ...+ \frac{(x-x_0)(x-x_1)...(x-x_{n-1})}{(x_0 - x_1)(x_0 - x_2)...(x_0 - x_n)}$$

especially the 1 at the very front, where does it come from?

EDITED:

I see the number 1 is probably from this property of the Lagrange basis function (here's where I found it if anyone is interested in learning):

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

and if we arrange everything to the left, it would be:

$$l_0(x) = 1 - \sum_{i=1}^n l_i(x)$$

If I am starting correctly, the next step would be expanding $\sum_{i=1}^nl_i(x)$ which I am looking for more enlightenment..

Answer 1

Hint: Prove that $g(x_j) = \delta_{0j}$, where $$g(x) = 1 + \frac{x-x_0}{x_0 - x_1} + \frac{(x-x_0)(x-x_1)}{(x_0 - x_1)(x_0 - x_2)} + \cdots+ \frac{(x-x_0)(x-x_1)\cdots(x-x_{n-1})}{(x_0 - x_1)(x_0 - x_2)\cdots(x_0 - x_n)}$$ Then $l_0-g$ is a polynomial of degree at most $n$ with $n+1$ zeros, and so must be the zero polynomial.

I assume that $l_{ij}$ is the Lagrange basis function such that $l_i(x_j) = \delta_{ij}$.