Solutions

a) The hint is

$$\prod_{\substack{j=1\\ j\ne i}}^n (x-x_j)=0$$

when $x=x_j$, where $j\ne i$. When $x=x_i$, the expression becomes

$$\prod_{\substack{j=1\\ j\ne i}}^n (x_i-x_j),$$

which we can make equal to $1$ by dividing each factor by itself, which would give

$$f_i(x)=\prod_{\substack{j=1\\ j\ne i}}^n \frac{x-x_j}{x_i-x_j}.$$

b) Using the function from a), we can create

$$f(x)=a_1f_1(x)+a_2f_2(x)+\cdots+a_nf_n(x).$$

That is a polynomial function of degree $n-1$ since it is a sum of polynomials of degree $n-1$. Moreover, by definition of $f_i(x)$, we have that

$$\begin{align*} f(x_i) &= a_1f_1(x_i)+a_2f_2(x_i)+\cdots+a_nf_n(x_i) \\ &= a_1\cdot 0+a_2\cdot 0+\cdots+a_i\cdot 1+\cdots+a_n\cdot 0 \\ &= a_i. \end{align*}$$

So the conditions on $f$ are satisfied.