We proceed by induction on $r$. Let $r=1$. Applying $\beta_{0}+1$ successive integrations by parts, the integral computes as

$$\sum_{j=0}^{\beta_{0}}\left[\frac{(1-w_{1})^{j}}{j!}f^{(j)}(w_{1})\right]_{0}^{1}=f(1)-\sum_{j=0}^{\beta_{0}}\frac{f^{(j)}(0)}{j!},$$

as stated. Let $r\geq 2$. Applying the case $r-1$ to the function $f(z)=\frac{z^{|\beta|+r-1}}{(|\beta|+r-1)!}$,

$$\frac{1}{\beta_{0}!\dots\beta_{r-1}!}\int_{\Delta_{r-1}}w_{0}^{\beta_{0}}w_{1}^{\beta_{1}}\dots w_{r-1}^{\beta_{r-1}}\,\text{\rm d}w_{1}\wedge\dots\wedge\,\text{\rm d}w_{r-1}=\frac{1}{(|\beta|+r-1)!}$$

and, after rescaling,

$$\frac{1}{\beta_{0}!\dots\beta_{r-1}!}\int_{(1-w_{r})\Delta_{r-1}}w_{0}^{\beta_{0}}w_{1}^{\beta_{1}}\dots w_{r-1}^{\beta_{r-1}}\,\text{\rm d}w_{1}\wedge\dots\wedge\,\text{\rm d}w_{r-1}=\frac{(1-w_{r})^{|\beta|+r-1}}{(|\beta|+r-1)!}.$$

Therefore, the left-hand side of the equality to be proved reduces to

$$\frac{1}{(|\beta|+r-1)!}\int_{0}^{1}(1-w_{r})^{|\beta|+r-1}f^{(|\beta|+r)}(w_{r})\,\text{\rm d}w_{r}.$$

Applying the case $r=1$ and index $|\beta|+r-1\in\mathbb{N}$, we find that this integral equals $f(1)-\sum_{j=0}^{|\beta|+r-1}f^{(j)}(0)/j!$, which concludes the proof. ∎