n-dimensional Lebesgue measure of a set in terms of gamma function Im having troubles proving by induction that the measure of the set $C_{n,p} = \{x \in \mathbb{R}^n : x_1^p + x_2^p + \dots + x_n^p \leq 1, x_i \geq 0  \}$ is equal to $\frac{\left(\Gamma \left(\frac{1}{p} + 1 \right)\right)^n}{\Gamma \left(\frac{n}{p}+1\right)} $ but I can't figure out the inductive step. Any ideas?
 A: I'll describe the elementary proof I could figure out. The flavour is not all that inductive - I instead develop an explicit relation between $C_{n,p}$ and $C_{n-1,p}$ that can be easily solved. I guess in principle you could treat it inductively, but it doesn't seem so natural.

There are two tricks. The first is just a question of scaling. Let $V_{n,p}(r)$ be the $n$-measure of the set $\{ \sum x_i^p \le r^p, x_i \ge 0\}$. Then $V_{n,p}(r) = C_{n,p} r^n.$ This is fundamental to the behaviour of the Lebesgue measure, but consider the following proof if you like \begin{align} V_{n,p}(r) &= \int \mathbf{1}\{ \sum x_i^p \le r^p, x_i \ge 0\} \mathrm{d}x \\ &= \int \mathbf{1}\{ \sum (x_i/r)^p \le 1, x_i/r \ge 0\} \mathrm{d}x \\ &= \int \mathbf{1}\{\sum y_i^p \le 1, y_i \ge 0\} r^n \mathrm{d}y,\end{align} using the substitution $y = x/r,$ and where $\mathbf{1}\{ \cdot \}$ is the indicator function - its $1$ when the relation holds, $0$ when not. 
The next trick is establishing a recursion via Fubini's theorem $$C_{n,p} = \int_{ \sum x_i^p \le 1, x_i \ge 0} \mathrm{d}x = \int_{x_n = 0}^1 \int_{\sum_{i= 1}^{n-1}x_i^p \le 1 - x_n^p, x_i \ge 0} \,\mathrm{d}x_1^{n-1} \mathrm{d}x_n = \int_{u = 0}^1 C_{n-1, p} (1-u^p)^{n-1/p} \mathrm{d}u, $$ where $\mathrm{d}x_1^{n-1}$ is the $n-1$-volume element on the first $n-1$ coordinates, and the final equality follows since the inner integral is just $V_{n-1,p}( (1 - x_n^p)^{1/p}).$
Now it's just a question of computing the integral, and unrolling the recursion, utilizing that $C_{1,p} = 1$ for every $p$. We compute this integral by reducing to the beta function - using the substitution $u = t^{1/p},$ $$\int_0^1 (1-u^p)^{n-1/p} \mathrm{d} u = \int_0^1  (1-t)^{n-1/p} \frac{t^{1/p-1}}{p}\mathrm{d}t = B((n-1/p) + 1, 1/p)/p\\ = \frac{\Gamma((n-1/p) + 1) \Gamma(1/p)}{p \Gamma(n/p + 1)} = \frac{\Gamma(1 + \frac{n-1}{p}) \Gamma(1 + \frac{1}{p})}{\Gamma(1 + \frac{n}{p})} $$
All equalities after the first are standard properties of the Beta function and the Gamma function. It should be straightforward to conclude from here.
