Integrate function over 3-simplex I am working on the following exercise, which is not homework:

Let $T$ be the simplex $\left\{(x, y, z) \in \mathbb{R} \mid x, y, z \ge 0, x + y + z \le 1\right\}$. Compute $\int_T xyz(1-x-y-z)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z$.

It is certainly possible to rewrite the integral as a triple integral
$$\int_0^1\int_0^{1-z}\int_0^{1-y-z} xyz(1-x-y-z)\,\mathrm{d}x\mathrm{d}y\mathrm{d}z,$$
but this leads to many additive terms and is not very pleasant to evaluate. Are there any tricks to make computation easier?
Edit:
Thank you very much, coffeemath! With the change of variables you proposed, I could finally compute the integral. Indeed, the result seems to be $1/((2n+1)!)$ for a n-simplex. Not easy to see, though…
 A: With a change of variables the integral looks simpler, and when evaluated gives somewhat simple intermediate results. Let 
$$u=x+y+z,\\ v=x+y,\\w=x,$$ so that the norm of the Jacobian of the inverse coordinate change is $1$. The region of integration is now $0 \le w \le v \le u \le 1.$ And in these variables the integrand is $w(v-w)(u-v)(1-u).$ So as an iterated integral it is
$$\int_0^1 \int_0^u \int_0^v w(v-w)(u-v)(1-u) \ dw \ dv \ du.$$
I'm not saying there isn't anything "messy" in the steps after this, only that they are a bit simpler in complexity compared to the original integral set up as an iterated integral as in the posted question.
Actually the intermediate integrals aren't that messy. After integrating over $w$ it is $(1-u)(u-v)v^3/6,$ and after the next integration over $v$ it is $(1-u)u^5/120,$ finally the last integration over $u$ brings it to $1/5040,$ which is $1/7!$ which makes me think there may be a pattern for more variables.
A: Here is a way to explain the $1/7!$ by, rather than doing the integrals, turn each factor into another integral.  You then have an integral of 1 over part of the 7-cube, with variables in a specified order.  By symmetry, that is $1/7!$ of the integral of 1 over the whole cube.
From @coffeemath,
$$\int_0^1du\int_{v\lt u}dv\int_{w\lt v}dw w(v-w)(u-v)(1-u)\\
=\int_{w\lt v\lt u\lt1}\int_0^w dt\int_w^vds\int_v^udr\int_u^1dq\\
=\int_{0\lt t\lt w\lt s\lt v\lt r\lt u\lt q\lt 1}1dtdwdsdvdrdudq\\
=\frac1{7!}\int_{[0,1]^7}1 dV =1/5040$$
