So the problem is this;
For Lagrange Interpolation on the nodes $x_0 < x_1 < \ldots < x_{n-1} < x_n$ of the data ${(x_i, f(x_i))}^{n}_{i=0}$ the interpolating polynomial is $\sum f(x_i)L_i(x)$, where $L_i(x) = \frac{(x-x_0)\ldots(x-x_{i-1})(x-x_{i+1})\ldots(x_i-x_n)}{(x_i-x_0)\ldots(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}.$ Prove that $L_0(x) = 1 + \frac{(x-x_0)}{(x_0-x_1)}+\frac{(x-x_0)(x-x_1)}{(x_0-x_1)(x_0-x_2)}+\ldots+\frac{(x-x_0)(x-x_1)\ldots(x-x_{n-1})}{(x_0-x_1)(x_0-x_2)...(x_0-x_n)}$ And state the general result for $L_i(x)$.
I was able to prove the first part without issue. \begin{align} & = 1 + \frac{(x-x_0)}{(x_0-x_1)}+ \frac{(x-x_0)(x-x_1)}{(x_{0}-x_1)(x_{0}-x_2)} + \ldots + \frac{(x-x_0)(x-x_1)\ldots(x-x_k)}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})}\\ & = \frac{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \frac{(x-x_0)(x_0-x_2)\ldots(x_0-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \frac{(x-x_0)(x-x_1)\ldots(x-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \ldots + \frac{(x-x_0)(x-x_1)\ldots(x-x_k)}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})}\\ & = \frac{(x_0-x_2)(x-x_1)\ldots(x_0-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \frac{(x-x_0)(x-x_1)\ldots(x-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \ldots + \frac{(x-x_0)(x-x_1)\ldots(x-x_k)}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})}\\ & = \frac{(x-x_1)(x-x_2)\ldots(x_{0}-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} + \frac{(x-x_0)(x-x_1)\ldots(x-x_k)}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})}\\ & = \frac{(x-x_1)(x-x_2)\ldots(x-x_{k+1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_{k+1})} \end{align}
So the inductive hypothesis holds for $L_0(x)$ on nodes $x_0< \ldots <x_k<x_{k+1}$.
I did some experimenting and expanded a couple polynomials $L_i$...
$L_0(x) = 1+\frac{(x-x_0)}{(x_0-x_1)}+\frac{(x-x_0)(x-x_1)}{(x_0-x_1)(x_0-x_2)}+\ldots+\frac{(x-x_0)(x-x_1)\ldots(x-x_{n-1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_n)}$
$L_1(x) = \frac{(x-x_1)}{(x_1-x_0)}+\frac{(x-x_0)(x-x_1)}{(x_0-x_1)(x_1-x_2)}+\ldots+\frac{(x-x_0)(x-x_1)\ldots(x-x_{n-1})}{(x_0-x_1)(x_0-x_2)\ldots(x_0-x_n)}$
$L_2(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}+\ldots+\frac{(x-x_0)(x-x_1)(x-x_2)\ldots(x-x_{n-1})}{(x_2-x_0)(x_2-x_1)\ldots(x_2-x_n)}$
Thank you; Bray.
I think, if I am asked for an analogue for the result, I would answer,
$$L_i(x) = 1 + \frac{x - x_i}{x_i - x_{i+1}} + \frac{(x - x_i)(x - x_{i+1})}{(x_i - x_{i+1})(x_i - x_{i+2})} + \ldots + \frac{(x - x_i)(x - x_{i+1})\ldots(x - x_{n+i-1})}{(x_i - x_{i+1})(x_i - x_{i+2})\ldots(x_i - x_{n+i})}$$
where $x_{n+j} = x_j$ for $j \geq 1$.
Also, the proof for the first part is easily done by checking R.H.S. fulfils
$$x_j \mapsto \delta_{0,j}$$
and such a polynomial with degree not greater than $n$ is unique.