Barycentric weight is defined as :
$$w_j := \prod_{\substack{k=0\\k \ne j} }^{n} (x_j - x_k)^{-1}.$$
For any $n$, let $x_j = j$, $j = 0,\dots , n$. Show that
$$w_j = \frac{(-1)^{n-j}}{j!(n-j)!}.$$
The only thing that I got is
$$\biggl( j(j-1)(j-2)...(j-(j-1))(j-(j+1))...(j-n)\biggr)^{-1}.$$
Let's just say this is an exercise in counting. Remember that $j$ is any integer from the index set $\{0,1,\dots,n\}$. Let us examine the two end cases $w_0$ and $w_n$: \begin{align*} w_0 & = \frac{1}{(0-1)(0-2)\dots(0-n)} = \frac{(-1)^n}{1\cdot 2\cdot\dots\cdot n} = \frac{(-1)^n}{n!} = \frac{(-1)^{n-0}}{0!(n-0)!} \\ w_n & = \frac{1}{(n-0)(n-1)\dots(n-(n-1))} = \frac{1}{n\cdot (n-1)\cdot \dots \cdot 1} = \frac{1}{n!} = \frac{(-1)^{n-n}}{n!(n-n)!}. \end{align*} Now suppose $j$ is some integer between $0$ and $n$. We write the barycentric weight as follows: \begin{align*} w_j = \left(\prod_{k=0, k\neq j}^n (j - k)\right)^{-1} = \left(\prod_{k=0}^{j-1} (j-k)\prod_{k=j+1}^n (j-k)\right)^{-1}. \end{align*} For the first product, performing a change of variable $y = j-k$ yields $$ \prod_{k=0}^{j-1} (j-k) = \prod_{y=1}^j y = j!. $$ For the second product, performing a change of variable $z = k-j$ yields $$ \prod_{k=j+1}^n (j-k) = \prod_{z=1}^{n-j} (-z) = (-1)^{n-j}(n-j)!. $$ Combining everything, we thus obtain $$ w_j = \frac{1}{j!(-1)^{n-j}(n-j)!} = \frac{(-1)^{n-j}}{j!(n-j)!} \ \ \textrm{ for every $j=0,1,\dots,n$}, $$ as desired.