Is there an easy way to compute the size (Lebesgue measure) of the set $$S_n(a):=\{(x_1,\ldots,x_n)\in\mathbb{R}^n:x_1+\cdots+x_n<a\text{, and }x_i>0\}.$$ Using integration I computed that $$m(S_n(a))=\frac{a^n}{n!}.$$ But the computation is tedious, and I was wondering if there are more concise/direct ways of proving it.
Computation: $$ \begin{align} m(S_n(a)) &= \int_0^a\int_0^{a-x_1}\cdots\int_0^{a-x_1-\cdots-x_{n-1}}dx_n\cdots dx_1 \\ &= \int_0^a\int_0^{a-x_1}\cdots\int_0^{a-x_1-\cdots-x_{n-2}}(a-x_1-\cdots-x_{n-1})dx_{n-1}\cdots dx_1 \\ &= \int_0^a\int_0^{a-x_1}\cdots\int_0^{a-x_1-\cdots-x_{n-3}}\frac{-1}{2}(a-x_1-\cdots-x_{n-1})^2\big|_0^{a-x_1-\cdots-x_{n-2}}dx_{n-2}\cdots dx_1\\ &=\int_0^a\int_0^{a-x_1}\cdots\int_0^{a-x_1-\cdots-x_{n-3}}\frac{1}{2}(a-x_1-\cdots-x_{n-2})^2dx_{n-2}\cdots dx_1\\ \end{align} $$ and so on...
