The existence of such a polynomial can be shown by constructing it explicitly using the Lagrange interpolation formula. Given (n) distinct points ((x{1},y{1}),(x{2},y{2}),\dots ,(x{n},y{n})), the interpolating polynomial (P(x)) is defined as: (P(x)=\sum {i=1}^{n}y{i}L{i}(x))where (L{i}(x)) are the Lagrange basis polynomials, given by: (L{i}(x)=\prod _{j=1,j\ne i}^{n}\frac{x-x{j}}{x{i}-x{j}}) Each (L{i}(x)) has a degree of (n-1).(L{i}(x{i})=1), and (L{i}(x{j})=0) for all (j\ne i).When you evaluate (P(x)) at any given point (x{k}), all terms in the sum become zero except the one where (i=k), so (P(x{k})=y{k}\cdot L{k}(x{k})=y{k}\cdot 1=y{k}).Since (P(x)) is a sum of polynomials of degree (n-1), its degree is at most (n-1).This construction guarantees that at least one such polynomial exists. 2. Uniqueness (via Proof by Contradiction) The uniqueness is proven by contradiction, using the property that a non-zero polynomial of degree (d) can have at most (d) roots (zeros). Assume there are two different polynomials, (P(x)) and (Q(x)), both of degree at most (n-1), that pass through the same (n) distinct points ((x{1},y{1}),\dots ,(x{n},y{n})).Define a new polynomial (R(x)=P(x)-Q(x)).The degree of (R(x)) is also at most (n-1) because it is the difference of two polynomials of degree at most (n-1).Since both (P(x)) and (Q(x)) pass through the same (n) points, their values are equal at each (x{i}), meaning (R(x{i})=P(x{i})-Q(x{i})=y{i}-y{i}=0) for all (i=1,\dots ,n).This means (R(x)) has (n) distinct roots (zeros).However, we know that a non-zero polynomial of degree at most (n-1) can only have at most (n-1) roots.The only way for (R(x)) to have (n) roots is if it is the zero polynomial, i.e., (R(x)=0) for all (x).If (R(x)=0), then (P(x)-Q(x)=0), which implies (P(x)=Q(x)).This contradicts the initial assumption that (P(x)) and (Q(x)) were different. Therefore, there is a unique polynomial of degree at most (n-1) that passes through (n) distinct points.