= Solution
The <product rule> applied at the <interpolation node> $x_k$ gives
$$
w'(x_k)=\prod_{j\ne k}(x_k-x_j)\ne0,\qquad
\ell_k(x)=\frac{w(x)}{(x-x_k)w'(x_k)}=w(x)\frac{w_k}{x-x_k}.
$$
Thus <Lagrange interpolation> gives, away from the <interpolation nodes>,
$$
p_n(x)=w(x)\sum_{k=0}^n\frac{w_k f_k}{x-x_k}.
$$
Apply the same <Lagrange interpolation> formula to the <constant polynomial> one. Since its <interpolant> is itself,
$$
1=\sum_{k=0}^n\ell_k(x)=w(x)\sum_{k=0}^n\frac{w_k}{x-x_k}.
$$
Dividing the preceding two identities proves the <barycentric interpolation formula>
$$
\boxed{p_n(x)=\frac{\displaystyle\sum_{k=0}^n w_k f_k/(x-x_k)}{\displaystyle\sum_{k=0}^n w_k/(x-x_k)}.}
$$
The denominator is exactly $1/w(x)$, so is nonzero away from the <interpolation nodes>. At $x=x_j$, multiply numerator and denominator by $x-x_j$ and take the <limit>: their values tend to $w_jf_j$ and $w_j$, respectively. The <removable singularity> therefore has value $f_j$; in numerical evaluation return the stored value directly at an <interpolation node>.
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