It is known that the solid angle in a flat space of $d$ dimensions ($d = 2 n$ or $d = 2 n + 1$) is given by these formulae: \begin{align}\tag{1} \Omega_{2 n} &= \frac{1}{(n - 1)!} \, 2 \pi^n, \qquad &\Omega_{2 n + 1} &= \frac{2^{2 n} \, n!}{(2 n)!} \, 2 \pi^n. \end{align} For examples: $\Omega_1 = 2$, $\Omega_2 = 2 \pi$, $\Omega_3 = 4 \pi$, $\Omega_4 = 2 \pi^2$. In some papers, it is described as the volume $\mathrm{Vol}(\mathbb{S}^{d−1})$ of the $d - 1$ unit-sphere (is there a difference?).
Instead of calculating a volume of some object (the unit sphere), suppose an observer is in free fall in a $D$ dimensional spacetime, so space appears to be flat locally (space around the observer has $d = D - 1$ dimensions). The observer wants to calculate the solid angle all around himslef, by looking in every orientations. He should get (1). What would be the simplest way in deriving these expressions?
Let $\mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and $$ r\equiv\sqrt{x_1^2+\cdots+x_N^2}. $$ Define $$ \Omega(N) = \frac{f(N)}{g(N)} \tag{1} $$ with $$ f(N)\equiv \int d^N x\ \exp(-\mathbf{x}^2) \tag{2} $$ and $$ g(N)\equiv \int_0^\infty dr\ r^{N-1} \exp(-r^2). \tag{3} $$ The definition (1) implies that $\Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is $$ f(N)\equiv \left(\int dx\ e^{-x^2}\right)^N = \pi^{N/2}. \tag{4} $$ To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use \begin{align} g(N) &= \left(-\frac{d}{da}\right)^{(N-1)/2} \left.\int_0^\infty dr\ e^{-ar^2}\right|_{a=1} \\ &= \left.\left(-\frac{d}{da}\right)^{(N-1)/2} \sqrt{\frac{\pi}{4a}}\right|_{a=1}. \tag{5} \end{align} When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use \begin{align} g(N) &= \left.\left(-\frac{d}{da}\right)^{(N-2)/2} \int_0^\infty dr\ r\,e^{-ar^2}\right|_{a=1} \\ &= \left.\left(-\frac{d}{da}\right)^{(N-2)/2} \frac{1}{2a}\right|_{a=1}. \tag{6} \end{align}