Let $$E = \{(x_1,\ldots,x_n) \mid x_1,\dots,x_n>0, x_1+\cdots+x_n<1 \} , p_1,\dots,p_n>0$$ and $f:[0,1]\to \mathbb{R}$ be continuously differentiable. Prove $$\int \cdots \int_E f(x_1+ \cdots +x_n) x_1^{p_1-1}...x_n^{p_n-1}\, dx_1\cdots dx_n=\ \frac{\Gamma(p_1)...\Gamma(p_n)}{\Gamma(p_1+...+p_n)}\int_0^1 f(s)s^{p_1+...+p_n-1} \, ds$$
I'm not sure if it helps but earlier we proved that $$\int \cdots \int_E f(x_1+ \cdots +x_n) \, dx_1\cdots dx_n=\frac{1}{(n-1)!}\int_0^1 f(s)s^{n-1} \, ds$$ using induction and the following change of basis:
$$\begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ x_n \end{bmatrix} \mapsto \begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ x_1+\cdots + x_n \end{bmatrix} = \begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ s \end{bmatrix} $$
I also believe we should use the following identity:
$\int \cdots \int_E x_1^{p_1-1}...x_n^{p_n-1}\, dx_1\cdots dx_n= \frac{\Gamma(p_1)...\Gamma(p_n)}{\Gamma(p_1+...+p_n+1)}$
Any help?
I try a solution, somewhat complicated.
1) Let $F=[0,+\infty[^n$. Put for $t>0$
$$h(t)=\int ..\int_Fx_1^{p_1-1}..x_n^{p_n-1}\exp(-t(x_1+..+x_n))dx_1..dx_n$$
Then we have easily $\displaystyle h(t)=\frac{\Gamma(p_1)..\Gamma(p_n)}{t^{p_1+..+p_n}}$
Now let $m$ an integer $\geq 0$; derivating $m$ times, simplifying by $(-1)^m$ and putting $t=1$, we get that $$\int ..\int_F x_1^{p_1-1}..x_n^{p_n-1}(x_1+..+x_n)^m\exp(-(x_1+..+x_n))dx_1..dx_n$$ is equal to $$A=\Gamma(p_1)..\Gamma(p_n)(p_1+..+p_n)(p_1+..+p_n+1)..(p_1+..+p_n+m-1)$$
We expand now $$(x_1+..+x_n)^m=\sum_{j_1+.;+j_n=m}\frac{m!}{j_1!..j_n!}x_1^{j_1}..x_n^{j_n}$$, we multiply by $x^{p_1-1}..x_n^{p_n-1}\exp(-(x_1+.;+x_n))$, and we integrate over $F$. We get $$B=\sum_{j_1+..+j_n=m}\frac{m!}{j_1!..j_n!}\Gamma(j_1+p_1)..\Gamma(j_n+p_n)$$ Hence we have $A=B$.
2) Now we return to the problem. Take $f(x)=x^m$. Then by your identity at the end of your question, we have if $$C=\int..\int_E (x_1+..+x_n)^mx_1^{p_1-1}..x_n^{p_n-1}dx_1dx_n$$ that $$C=\sum_{j_1+..+j_n=m}\frac{m!}{j_1!..j_n!}\frac{\Gamma(p_1+j_1)..\Gamma(p_n+j_n)}{\Gamma(p_1+..+p_n+m+1)}$$
And if $$D= \frac{\Gamma(p_1)..\Gamma(p_n)}{\Gamma(p_1+..+p_n)}\int_0^1s^{m+p_1+..+p_n-1}ds$$ we have. $$D=\frac{\Gamma(p_1)..\Gamma(p_n)}{\Gamma(p_1+..+p_n)(m+p_1+..+p_n)}$$
Now using $A=B$, it is easy to show that $C=D$, hence the formula is true for $f(x)=x^m$. Now it is clear that the formula is true for all polynomials $P\in \mathbb{R}[x]$.
3) To finish, note that for an continuous function $f$ on $|0,1]$, there exists by Weiertrass' s theorem, a sequence of polynomials $P_q$, that converge uniformly to $f$ on $[0,1]$, and this show finally that your formula is true for any function $f$, continuous on $[0,1]$.