Form multi-dimensional to one-dimensional: What's the general form of the integral? From the indetities below
$$
\int_0^{\infty}dn_1f(n_1)=\int_0^{\infty}f(n)dn\\
\int_0^{\infty}dn_1\int_0^{\infty}dn_2f(n_1+n_2)=\int_0^{\infty}nf(n)dn\\
\int_0^{\infty}dn_1\int_0^{\infty}dn_2\int_0^{\infty}dn_3f(n_1+n_2+n_3)=\frac{1}{2}\int_0^{\infty}n^2f(n)dn
$$
We can conclude that
$$
\Big(\prod_{k=1}^{N}\int_0^{\infty}dn_k\Big)f(\sum_{k=1}^Nn_k)=C(N)\int_0^{\infty}n^{N-1}f(n)dn.
$$
How can we prove the general integral identity and find the $C(N)$ meanwhile? 
EDIT: As @Mickep points out in the Answer below, $C(N)=\frac{1}{\Gamma(N)}$.

Denote by $Q$ the first hyperoctant in ${\mathbb R}^n$ and by $Q'$ the first hyperoctant in ${\mathbb R}^{n-1}$ (the coordinate plane $x_n=0$ in ${\mathbb R}^n$). Furthermore introduce the abbreviations
$$(x_1,x_2,\ldots, x_{n-1})=:x',\qquad x_1+x_2+\ldots + x_{n-1}=:s'\ .$$
Then we have
$$\eqalign{\int_Qf(x_1+x_2+\ldots+x_n){\rm d}x
&=\int_{Q'}\int_0^\infty f(s'+x_n)\>dx_n\ {\rm d}x'\cr 
&=\int_{Q'}\int_{s'}^\infty f(t)\>dt\ {\rm d}x'\cr
&=\int_{Q'}\int_0^\infty f(t)\>1_{t\geq s'}\>dt\ {\rm d}x'\cr
&=\int_0^\infty f(t)\left(\int_{Q'} 1_{s'\leq t}\>{\rm d}x'\right) dt  \cr}$$
by Fubini's theorem. 
Now it is well known that the body $S_t:=\bigl\{x'\in Q'\bigm| s'\leq t\bigr\}$ (an "orthonormal simplex")  has volume $${\rm vol}_{n-1}(S_t)={t^{n-1}\over (n-1)!}\ .$$
We therefore obtain
$$\int_Qf(x_1+x_2+\ldots+x_n){\rm d}x={1\over(n-1)!}\int_0^\infty t^{n-1}\>f(t)\ dt\ .$$
