Let $x \in \mathbb{S}^{n}$ be a point on the unit $n$-sphere with coordinates $\begin{cases} x_1 &= \cos(\theta_1).\\ x_2 &= \sin(\theta_1)\cos(\theta_2).\\ x_3 &= \sin(\theta_1)\sin(\theta_2)\cos(\theta_3).\\ \vdots &\\ x_n &= \left(\prod_{i=1}^{n-1}\sin(\theta_i)\right)\cos(\theta_n).\\ x_{n+1} &= \left(\prod_{i=1}^{n-1}\sin(\theta_i)\right)\sin(\theta_n). \end{cases}$ where $0 \leq \theta_i \leq 2\pi, i = 1,\dots,n$. Take it as granted that the determinant of the Jacobian of the general polar coordinates is $\det(J) = \prod_{j=2}^{n+1}\sin^{n + 1 - j}(\theta_{j-1})$ and that the differential element $dx$ is given by $dx = \prod_{j=2}^{n+1}\sin^{n + 1 - j}(\theta_{j-1})d\theta_{j-1}$.
What should I do to find an expression for a normalization constant $c(k)$ such that $c(k)\int_{x \in \mathbb{S}^n}e^{k\mu^T x}dx = 1$ where $\mu \in \mathbb{R}^{n+1}: ||\mu||= 1$ and $k \in \mathbb{R}_{>0}$ are constants? My gut feeling is integration by parts and a possible change of variables, but the antiderivative of the exponential function will have cosines as coefficient and the Jacobian determinant has only sines. I know that some sort of black magic will yield an analytic expression because according to Wikipedia the normalization constant should be $c(k) = \frac{k^{(n+1)/2 - 1}}{(2\pi)^{(n+1)/2}I_{(n+1)/2 - 1}}$, link to von Mises-Fisher.
First, an error in your definition of $n$-spherical coordinates. The ranges of the variables are $0 < \theta_i <\pi$ for $1 \le i\le n-1$ and $0 < \theta_n < 2\pi$
With that out of the way, remember that when doing the integral you can choose any coordinate system to do it in. In particular, choosing one such that $\mu\cdot x = \cos\theta_1$ will simplify the integral considerably. You end up with \begin{multline} \int_{S^n} e^{k\cdot x}d^n x = \int_0^\pi e^{k\cos\theta_1}\sin^{n-1}\theta_1d\theta_1\int_{S^{n-1}}d^{n-1}x = \frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}\int_{-1}^1e^{ku}(1-u^2)^{\frac{n}{2}-1}du \\= \frac{2\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}\left[\sqrt{\pi}\left(\frac{2}{k}\right)^{\frac{n-1}{2}}\Gamma\left(\frac{n}{2}\right)I_{\frac{n-1}{2}}(k)\right] = \frac{(2\pi)^{\frac{n+1}{2}}}{k^\frac{n-1}{2}}I_\frac{n-1}{2}(k) , \end{multline} which coincides with the result given in the link.