To interpolate a polynomial of degree $n$ using the Lagrange form,
$$p(x)=\sum_{i=0}^ny_iL_i(x)$$ with $$L_i(x)=\frac{\prod_{i\not=j}(x-x_j)}{\prod_{i\not=j}(x_i-x_j)}$$
How can I show that for $y_i=x^j_i$ for $j=0, 1, ..., n$ $$\sum_{i=0}^nx_i^jL_i(x)=x^j$$
I read through this question which is the same identity, but the answers were unclear. I tried proving by expanding, so $$p(x)=\sum_{i=0}^n x_i^j \frac{\prod_{i\ne j} (x - x_j)}{\prod_{i \ne j} (x_i - x_j)}=\sum_{i=0}^n x_i^j \frac{(x - x_0)(x - x_1)...(x - x_{i-1})(x - x_{i+1})...(x - x_n)} {(x_i - x_0)(x_i - x_1)...(x_i - x_{i-1})(x_i - x_{i+1})...(x_i - x_n)}$$ but since the top product has $x$ where the denominator product has $x_i$, nothing seems to cancel out... but if I let $x=x_k$ as suggested in the link would that mean $x_i=x_k$ as well, leaving $\sum x_k^j=x_k^j$?