Proof. $$

1.

Remember from Theorem A-10 that Lagrange’s polynomial interpolation is defined as:

$f(X)=\underset{j=0}{\overset{n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}\left(\prod _{\begin{array}{c}0\le k\le n\\ k\ne j\end{array}}{\displaystyle \frac{X-x_k}{x_j-x_k}}\cdot y_j\right)$

$$

If every $x$, $y$ values and coefficients of the polynomial $f(X)$ lie in a field, then ${(x_j-x_k)}^{-1}$ is guaranteed to exist within the same field, because every non-zero element of a field has an inverse, and $(x_j-x_k)$ is not zero (since $x_j\ne x_k$). This implies that $\underset{\begin{array}{c}0\le k\le n\\ k\ne j\end{array}}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{\displaystyle \frac{X-x_k}{x_j-x_k}}\cdot y_j$ also lies in the same field. Hence, a well-defined $f(X)$ is guaranteed to exist. And the degree of $f(X)$ is at most $n$, because its highest possible degree term is $X^n$.

2.

For uniqueness of $f(X)$, suppose another polynomial $g(X)$ of degree at most $n$ also passes through the same $n+1$ points. Let $h(X)=f(X)-g(X)$. Then the degree of $h(X)$ is at most $n$, and $h(x_i)=0$ for all $(n+1)$ distinct nodes $x_0,\dots ,x_n$, so $h$ has $(n+1)$ distinct roots. Over a field, a non-zero polynomial of degree $d$ has at most $d$ roots; a nonzero $h(X)$ of degree at most $n$ could have at most $n$ roots. Since $h(X)$ has $(n+1)$ roots, $h(X)$ must be the zero polynomial, i.e. $f(X)=g(X)$. Therefore $f(X)$ is unique.

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