Existence and uniqueness of the interpolating polynomial

Step 1 (existence). The Lagrange construction $P(x) = \sum_{i=0}^{n} y_i L_i(x)$ with $L_i(x) = \prod_{j\ne i} \dfrac{x-x_j}{x_i-x_j}$ already satisfies $P(x_k)=y_k$ for every $k$, since $L_i(x_k)$ equals $1$ when $i=k$ and $0$ otherwise, so the sum collapses to the single term $y_k \cdot 1 = y_k$. So at least one valid $P$ exists, and each $L_i$ has degree exactly $n$ (a product of $n$ linear factors), so $P$ has degree at most $n$.

Step 2 (suppose two solutions). Suppose $Q$ is any other polynomial of degree at most $n$ with $Q(x_i)=y_i$ for every $i$. Consider the difference $D(x) = P(x) - Q(x)$. Since both $P$ and $Q$ have degree at most $n$, so does $D$.

Step 3 (count the roots of the difference). For every node $x_i$, $D(x_i) = P(x_i)-Q(x_i) = y_i - y_i = 0$. Since there are $n+1$ distinct nodes $x_0, x_1, \dots, x_n$, $D$ has at least $n+1$ distinct roots.

Step 4 (force D to be identically zero). A nonzero polynomial of degree at most $n$ can have at most $n$ roots (each root contributes one linear factor, and a degree-$n$ polynomial cannot contain more than $n$ such factors). Since $D$ has $n+1$ roots, more than its degree allows unless $D$ is the zero polynomial, we conclude $D(x)\equiv0$, i.e. $Q=P$. So the interpolating polynomial found in Step 1 is the only one.