1. **Problem Statement:** Given polynomials $p_0(x), p_1(x), p_2(x), \ldots, p_n(x)$ where each $p_k(x)$ is a polynomial of degree $k$ for $k=0,1,2,\ldots,n$, prove that the set $\{p_0(x), p_1(x), \ldots, p_n(x)\}$ forms a basis of the vector space $P_n$ of all polynomials of degree at most $n$. 2. **Recall the definition of a basis:** A set of vectors (here polynomials) is a basis of a vector space if it satisfies two conditions: - The set is linearly independent. - The set spans the vector space. 3. **Dimension of $P_n$:** The vector space $P_n$ has dimension $n+1$ because any polynomial of degree at most $n$ can be written as $$a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$ with coefficients $a_i$ in the field. 4. **Linear independence:** Suppose there exist scalars $c_0, c_1, \ldots, c_n$ such that $$c_0 p_0(x) + c_1 p_1(x) + \cdots + c_n p_n(x) = 0$$ for all $x$. Since $p_k(x)$ has degree exactly $k$, the polynomial on the left is a polynomial of degree at most $n$. 5. **Degree argument:** The highest degree term comes from $c_n p_n(x)$ which has degree $n$ if $c_n \neq 0$. For the sum to be zero polynomial, the coefficient of $x^n$ must be zero, so $c_n = 0$. Similarly, by descending degree, all $c_k = 0$ for $k=n-1, n-2, \ldots, 0$. 6. **Conclusion on linear independence:** Only the trivial combination yields the zero polynomial, so the set is linearly independent. 7. **Spanning:** Since the set has $n+1$ polynomials and $\dim P_n = n+1$, and the set is linearly independent, it must span $P_n$. 8. **Final conclusion:** Therefore, $\{p_0(x), p_1(x), \ldots, p_n(x)\}$ is a basis of $P_n$.