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, $C(N)=\frac{1}{\Gamma(N)}$.
 A: 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
$$\int_Qf(x_1+x_2+\ldots+x_n){\rm d}x={1\over(n-1)!}\int_0^\infty t^{n-1}\>f(t)\ dt\ .$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\dd n_{1}\int_{0}^{\infty}\dd n_{2}\ldots
\int_{0}^{\infty}\dd n_{N - 1}\int_{0}^{\infty}\dd n_{N}\,
\fermi\pars{n_{1} + n_{2} + \cdots + n_{N}}}
\\[5mm]&=\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}\dd n_{N - 1}
\int_{n_{1}\ +\ n_{2}\ +\ \cdots\ +\ n_{N - 1}}^{\infty}\dd n_{N}\,\fermi\pars{n_{N}}
\\[5mm]&=\left.\int_{0}^{\infty}\dd n\,\fermi\pars{n}
\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}\dd n_{N - 1}\,
\right\vert_{\, n_{1}\ +\ n_{2}\ +\ \cdots\ +\ n_{N - 1}\,\,\,\, <\ n}
\\[5mm]&=\int_{0}^{\infty}\fermi\pars{n}\phi_{N}\pars{n}\,\dd n.\qquad
\pars{~\mbox{we rename the}\ n_{N}\ \mbox{variable:}\ n_{N}\ \mapsto\ n ~}
\end{align}

$\ds{\phi_{N}\pars{n}}$ is given by
  $\ds{\phi_{N}\pars{n}\equiv
\left.\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}\dd n_{N - 1}\,
\right\vert_{\, n_{1}\ +\ n_{2}\ +\ \cdots\ +\ n_{N - 1}\,\,\,\, <\ n}}$

\begin{align}
\phi_{N}\pars{n}&=
\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}
\Theta\pars{n - n_{1} - n_{2} - \cdots - n_{N - 1}}\,\dd n_{N - 1}
\end{align}

$\ds{\Theta\pars{x}}$ is the
  Heaviside Step Function
  which, indeed, is quite helpful. Below we use, in addition, the Dirac Delta Function:

\begin{align}
\phi'_{N}\pars{n}&=
\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}
\delta\pars{n - n_{1} - n_{2} - \cdots - n_{N - 1}}\dd n_{N - 1}
\\[5mm]&=\int_{0}^{\infty}\dd n_{1}
\int_{0}^{\infty}\dd n_{2}\ldots\int_{0}^{\infty}
\Theta\pars{n - n_{1} - n_{2} - \cdots - n_{N - 2}}\,\dd n_{N - 2}
=\phi_{N - 1}\pars{n}
\end{align}

Then, we get the recurrence:

\begin{align}
\phi_{N}''\pars{n}&=\phi'_{N - 1}\pars{n}=\phi_{N - 2}\pars{n}\,,\quad
\phi_{N}'''\pars{n}=\phi'_{N - 2}\pars{n}=\phi_{N - 3}\pars{n}
\end{align}

and, in general,

$$
\phi_{N}^{\rm\pars{N - 2}}\pars{n}=\phi_{2}\pars{n}
=\int_{0}^{\infty}\Theta\pars{n - n_{1}}\,\dd n_{1}=n
$$
which yields:
\begin{align}&
\phi_{N}^{\rm\pars{N - 3}}\pars{n}={n^{2} \over 2}\,,\qquad
\phi_{N}^{\rm\pars{N - 4}}\pars{n}={n^{3} \over 3 \times 2}\,,\qquad
\phi_{N}^{\rm\pars{N - 5}}\pars{n}={n^{4} \over 4 \times 3 \times 2}
\\[5mm]&\imp\qquad
\phi_{N}^{\rm\pars{k}}\pars{n}
={n^{N - k - 1} \over \pars{N - k - 1}!}\quad\imp\quad
\dsc{\phi_{N}\pars{n}=\phi_{N}^{\rm\pars{0}}\pars{n}
={n^{N - 1} \over \pars{N - 1}!}}
\end{align}

Then

\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\dd n_{1}\int_{0}^{\infty}\dd n_{2}\ldots
\int_{0}^{\infty}\dd n_{N - 1}\int_{0}^{\infty}\dd n_{N}\,
\fermi\pars{n_{1} + n_{2} + \cdots + n_{N}}}
\\[5mm]&=\color{#66f}{\large%
{1 \over \pars{N - 1}!}\int_{0}^{\infty}\fermi\pars{n}n^{N - 1}\,\dd n}
\end{align}
A: This only shows how to find the $C(N)$, given that the equality is true:
If you know that such an equality should hold, you only need to take one certain function. Let $f(x)=\exp(-x)$. Then you will find that the left-hand-side factors into $N$ integrals which are all equal to $1$. The integral $\int_0^{+\infty} n^{N-1}\exp(-n)\,dn=\Gamma(N)$. Thus, $C(N)=\frac{1}{\Gamma(N)}=\frac{1}{(N-1)!}$.
