Recurrence for expected length of Gaussian vector Let $g_k \sim N(0, I_{k \times k})$ be a a standard $k$-dimensional Gaussian vector.
Denote by $\|g\|$ the $2$-norm of $g$. By explicit integration, it is not hard to see that
$$
\mathbb E \|g_k\| = \frac{\sqrt 2 \Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac k 2\right)}\,,
$$
where $\Gamma$ is the Gamma function.
In particular, the above expression implies
\begin{equation*}
\mathbb E \|g_k\|\mathbb E\|g_{k+1}\| = k\,.
\end{equation*}
This formula gives a nice recurrence for the expected length of a standard Gaussian. The recurrence is so nice that I'd like to see a slick proof of this fact, if one exists.
Question: Is there an elegant proof of the recurrence $\mathbb E \|g_k\|\mathbb E\|g_{k+1}\| = k$, one that involves no explicit integration?
 A: In dimension $k$ there is a constant $C_k$ such that $$C_k \int_0^{\infty}r^{k-1}e^{-r^2/2}dr = 1.$$ It is not necessary to compute these constants as we will see. Now the expression for the product of expectations in radial form is $$\mathbb{E} \left(\lVert g_k \rVert\right)\mathbb{E} \left(\lVert g_{k+1} \rVert\right)=C_k\int_0^{\infty}r^k e^{-r^2/2}dr \cdot C_{k+1}\int_0^{\infty}r^{k+1} e^{-r^2/2}dr=$$ $$C_k\int_0^{\infty} r^2 r^{k-1} e^{-r^2/2} dr = \mathbb{E} \left(\lVert g_k \rVert^2 \right) = \mathbb{E} \left(X_1^2+\ldots+X_k^2\right)=k \mathbb{E} \left(X_1^2\right)=k.$$ Here $X_1,\ldots,X_k$ are iid standard normal variables.
A: A "slick" proof can be constructed using polar coordinates for the $k$-dimensional integral. With $r=\|g_k\|$ we define, for any function $f(r)$, 
$$ \langle f(r) \rangle = \int_0^\infty {\rm e}^{-\frac{r^2}2}f(r)\ {\rm d}r \ ,$$
so that 
$$ \mathbb E f(r) = \langle r^{k-1}f(r) \rangle \ .$$
We then have 
$$\mathbb E \|g_k\| = \frac{\langle r^k \rangle}{\langle r^{k-1} \rangle} \quad {\rm and} \quad \mathbb E \|g_k\|^2 = \frac{\langle r^{k+1} \rangle}{\langle r^k \rangle} \ ,$$
and as a consequence 
$$ \mathbb E \|g_k\| \ \mathbb E \|g_{k+1}\| 
= \frac{\langle r^k \rangle}{\langle r^{k-1} \rangle} \frac{\langle r^{k+1} \rangle}{\langle r^k \rangle}
= \mathbb E \|g_k\|^2 = k \ .$$
