To prove this result it suffices to compute the two integrals appearing in Lemma 8.22. However

$$L(y)=a_{0}+\sum_{\ell=1}^{r}(a_{\ell}-a_{0})y_{\ell}=a_{0}y_{0}+\dots+a_{r}y_{r},$$

with $y_{0}=1-y_{1}-\dots-y_{r}$, and therefore

$$L(y)^{n}=\sum_{\stackrel{{\scriptstyle\alpha\in\mathbb{N}^{r+1}}}{{\scriptscriptstyle|\alpha|=n}}}\binom{n}{\alpha_{0},\dots,\alpha_{r}}\prod_{\ell=0}^{r}(a_{\ell}y_{\ell})^{\alpha_{\ell}}$$

and similarly for $L(y)^{n+1}$. Now, Corollary 7.19 gives :

	$\displaystyle\int_{\Delta^{r}}y_{0}^{\alpha_{0}}y_{1}^{\alpha_{1}}\dots y_{r}^{\alpha_{r}}\,\text{\rm d}y$	$\displaystyle=\frac{\alpha_{0}!\dots\alpha_{r}!}{(|\alpha|+r)!},$
	$\displaystyle\int_{\Delta^{r}}y_{0}^{\alpha_{0}}y_{1}^{\alpha_{1}}\dots y_{r}^{\alpha_{r}}\log(y_{j})\,\text{\rm d}y$	$\displaystyle=-\frac{\alpha_{0}!\dots\alpha_{r}!}{(|\alpha|+r)!}\sum_{\ell=\alpha_{j}+1}^{|\alpha|+r}\frac{1}{\ell},$

which, combined with the above expression for $L(y)^{n}$ and $L(y)^{n+1}$, gives

	$\displaystyle\int_{\Delta^{r}}L(y)^{n}dy$	$\displaystyle=\sum_{\stackrel{{\scriptstyle\alpha\in\mathbb{N}^{r+1}}}{{\scriptscriptstyle|\alpha|=n}}}\frac{n!}{(n+r)!}\prod_{\ell=0}^{r}a_{\ell}^{\alpha_{\ell}}=\sum_{\stackrel{{\scriptstyle i_{0},\dots,i_{r}\in\mathbb{N}}}{{\scriptscriptstyle i_{0}+\dots+i_{r}=n}}}\prod_{\ell=0}^{r}a_{\ell}^{i_{\ell}}$
	$\displaystyle\int_{\Delta^{r}}L(y)^{n+1}dy$	$\displaystyle=\sum_{\stackrel{{\scriptstyle\alpha\in\mathbb{N}^{r+1}}}{{\scriptscriptstyle|\alpha|=n+1}}}\frac{(n+1)!}{(n+1+r)!}\prod_{\ell=0}^{r}a_{\ell}^{\alpha_{\ell}}=\sum_{\stackrel{{\scriptstyle i_{0},\dots,i_{r}\in\mathbb{N}}}{{\scriptscriptstyle i_{0}+\dots+i_{r}=n+1}}}\prod_{\ell=0}^{r}a_{\ell}^{i_{\ell}}$
	$\displaystyle\int_{\Delta^{r}}L(y)^{n}\varepsilon_{r}(y)dy$	$\displaystyle=-\sum_{m=0}^{r}\sum_{\stackrel{{\scriptstyle\alpha\in\mathbb{N}^{r+1}}}{{\scriptscriptstyle|\alpha|=n}}}\frac{n!(\alpha_{m}+1)}{(n+1+r)!}\bigg(\prod_{\ell=0}^{r}a_{\ell}^{\alpha_{\ell}}\bigg)\sum_{\ell=\alpha_{m}+2}^{n+1+r}\frac{1}{\ell}$
		$\displaystyle=-\frac{n!}{(n+1+r)!}\sum_{\stackrel{{\scriptstyle i_{0},\dots,i_{r}\in\mathbb{N}}}{{\scriptscriptstyle i_{0}+\dots+i_{r}=n}}}\bigg(\prod_{\ell=0}^{r}a_{\ell}^{i_{\ell}}\bigg)\sum_{m=0}^{r}(i_{m}+1)\sum_{\ell=i_{m}+2}^{n+1+r}\frac{1}{\ell}$
		$\displaystyle=-\frac{2\,n!}{(n+1+r)!}\sum_{\stackrel{{\scriptstyle i_{0},\dots,i_{r}\in\mathbb{N}}}{{\scriptscriptstyle i_{0}+\dots+i_{r}=n}}}\bigg(\prod_{\ell=0}^{r}a_{\ell}^{i_{\ell}}\bigg)A_{n,r}(i_{0},\dots,i_{r}).$

The statement follows from these expressions together with Lemma 8.22. ∎