Step 3. The polynomial ${\prod }_{i<j}(x_i-x_j)$ is homogeneous of degree $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\text{.}$

		Proof.
	Each monomial in the expansion of ${\prod }_{i<j}(x_i-x_j)$ is obtained by choosing a factor of either $x_i$ or $x_j$ from each factor $(x_i-x_j)$ for $1\le i<j\le n\text{.}$ Therefore, this polynomial is homogeneous of total degree $\left|\left\{(i,j)\hspace{0.17em}\right|\hspace{0.17em}1\le i<j\le n\}\right|=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\text{.}$ $\square$

Step 4. The coefficient of $x^{\lambda +\delta }$ in $a_{\lambda +\delta }\left(x\right)$ is one.

		Proof.
	Observe that $\lambda +\delta$ has distinct parts, so $w{x}^{\lambda +\delta }=x^{\lambda +\delta }$ implies $w=1\text{.}$ Then $$a_{\lambda +\delta }\left(x\right){|}_{x^{\lambda +\delta }}=\sum _{w\in W}\left(\epsilon \left(w\right)w{x}^{\lambda +\delta }\right){|}_{x^{\lambda +\delta }}=1\text{.}$$ $\square$

Step 5. $\left({\prod }_{i<j}(x_i-x_j)\right){|}_{x^{\delta }}=1\text{.}$

		Proof.
	To obtain the monomial $x^{\delta }=x_1^{n-1}+x_2^{n-2}+\cdots +x_{n-1}$ as a term in the expansion of ${\prod }_{i<j}(x_i-x_j),$ it is necessary to choose a factor of $x_1$ from all factors of the form $(x_1-x_j),$ $j>1\text{.}$ The only source of factors $x_2$ then are the factors of the form $(x_2-x_j),$ $j>2,$ from each of which we must choose $x_2\text{.}$ We proceed in this manner and are forced to choose all factors of $x_i$ from the terms $(x_i-x_j)$ where $i<j$ for $1\le i\le n-1\text{.}$ Hence, there is only one monomial equal to $x^{\delta }$ in the expansion of this product. $\square$

$\square$

The polynomials $a_{\lambda +\delta }\left(x\right)$ are a basis for the space of alternating symmetric functions, hence ${\prod }_{i<j}(x_i-x_j)={\sum }_{\lambda ⊢n}{c}_{\lambda +\delta }{a}_{\lambda +\delta }\left(x\right)$ for some $c_{\alpha }\in ℂ\text{.}$ The left hand side is homogeneous of degree $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ by step 3; however, only $a_{\delta }$ has this degree among the $a_{\lambda +\delta }$ by step one. Thus, ${\prod }_{i<j}(x_i-x_j)=c_{\delta }{a}_{\delta }\text{.}$ Comparing coefficients of $x^{\delta }$ using steps 4 (with $\lambda =0\text{)}$ and 5 yields that $c_{\delta }=1\text{.}$

$\square$

We roll up our sleeves and calculate. Let $\Delta =\left|\frac{1}{1-x_i{y}_j}\right|\text{.}$ Subtract the first row from all other rows; the $\left(i,j\right)$ entry of rows two through $n$ becomes $$\frac{1}{1-x_i{y}_j}-\frac{1}{1-x_1{y}_j}=\frac{(x_i-x_1)y_j}{(1-x_i{y}_j)(1-x_1{y}_j)}\text{.}$$ Hence, we may pull out a common factor of $(x_i-x_1)$ from row $i$ for $i\ge 2\text{.}$ Note that the product of these common factors may be written ${(-1)}^{n-1}{\prod }_{i=2}^n(x_1-x_i)\text{.}$ The determinant then becomes $$\Delta ={(-1)}^{n-1}\prod _{i=2}^n(x_1-x_i)\mid \begin{array}{cccc}\frac{1}{1-x_1{y}_1}& \frac{1}{1-x_1{y}_2}& \cdots & \frac{1}{1-x_1{y}_n}\\ \frac{y_1}{(1-x_2{y}_1)(1-x_1-y_1)}& \frac{y_2}{(1-x_2{y}_2)(1-x_1-y_2)}& \cdots & \frac{y_n}{(1-x_2{y}_n)(1-x_2-y_n)}\\ & ⋮\\ \frac{y_1}{(1-x_n{y}_1)(1-x_1-y_1)}& \frac{y_2}{(1-x_n{y}_2)(1-x_1-y_2)}& \cdots & \frac{y_n}{(1-x_n{y}_n)(1-x_1-y_n)}\end{array}\mid \text{.}$$ Extracting a common factor of ${(1-x_1{y}_j)}^{-1}$ from the $j\text{th}$ column for $1\le j\le n,$ we obtain $$\Delta ={(-1)}^{n-1}\prod _{i=2}^n(x_1-x_i)\prod _{j=1}^n\frac{1}{1-x_1{y}_j}\mid \begin{array}{cccc}1& 1& \cdots & 1\\ \frac{y_1}{1-x_2{y}_1}& \frac{y_2}{1-x_2{y}_2}& \cdots & \frac{y_n}{1-x_2{y}_n}\\ & ⋮\\ \frac{y_1}{1-x_n{y}_1}& \frac{y_2}{1-x_n{y}_2}& \cdots & \frac{y_n}{1-x_n{y}_n}\end{array}\mid \text{.}$$ Next subtract column one from each of the remaining columns. For rows $2\le i\le n,$ the $(i,j)$ entry is given by $$\frac{y_j}{1-x_i{y}_j}-\frac{y_1}{1-x_i{y}_1}=\frac{y_j-y_1}{(1-x_i{y}_j)(1-x_i{y}_1)}\text{.}$$ Hence we may extract a factor of $(-1)(y_1-y_j)$ from column $j$ for $2\le j\le n\text{.}$ Combining factors of ${(-1)}^{n-1},$ we obtain $$\begin{array}{ccc}\Delta & =& \prod _{i=2}^n(x_1-x_i)(y_1-y_i)\prod _{j=1}^n\frac{1}{1-x_1{y}_j}\mid \begin{array}{cccc}1& 0& \cdots & 0\\ \frac{y_1}{1-x_2{y}_1}& \frac{1}{(1-x_2{y}_2)(1-x_2{y}_1)}& \cdots & \frac{y_1}{(1-x_2{y}_n)(1-x_2{y}_1)}\\ & ⋮\\ \frac{y_1}{1-x_n{y}_1}& \frac{1}{(1-x_n{y}_2)(1-x_n{y}_1)}& \cdots & \frac{1}{(1-x_n{y}_n)(1-x_n{y}_1)}\end{array}\mid \\ & =& \prod _{i=2}^n(x_1-x_i)(y_1-y_i)\prod _{j=1}^n\frac{1}{1-x_1{y}_j}{\left|\frac{1}{(1-x_i{y}_j)(1-x_i{y}_1)}\right|}_{2\le i,j\le n}\end{array}$$ by expanding along the first row. We may pull out a factor of ${(1-x_i{y}_1)}^{-1}$ from each column $(2\le j\le n)\text{;}$ it follows from the inductive hypothesis for the variables $x_2,x_3,\dots x_n,y_2,y_3,\dots ,y_n$ that $$\begin{array}{ccc}\Delta & =& \prod _{i=1}^n(x_1-x_i)(y_1-y_i)\prod _{j=1}^n\frac{1}{1-x_1{y}_j}\frac{1}{1-x_j{y}_1}{\left|\frac{1}{(1-x_i{y}_j)}\right|}_{2\le i,j\le n}\\ & =& \prod _{i=1}^n(x_1-x_i)(y_1-y_i)\prod _{j=1}^n\frac{1}{1-x_1{y}_j}\frac{1}{1-x_j{y}_1}\left\{\prod _{2\le i<j\le n}(x_i-x_j)(y_i-y_j)\frac{1}{1-x_i{y}_j}\right\}\\ & =& a_{\delta }\left(x\right)a_{\delta }\left(y\right)\prod _{1\le i<j\le n}\frac{1}{1-x_i{y}_j}\end{array}$$

$\square$