Prove $\int_{\mathbb{R}^n}e^{-\max\{|x_1|,\ldots,|x_n|\}}dx=2^nn!$

There are $n$ cases, all of them mutually exclusive (except on a set of measure 0): $E_k=\{|x_k|=\max\{|x_j|:1\leq j\leq n\}$

One each $E_k$, by Fubini's theorem, the integral $$\int_{E_k}e^{-\max_{1\leq j\leq n}|x_j|}=\int_{\mathbb{R}}e^{-|x|}\Big(\int^{|x|}_{-|x|}\,dx_2\ldots\int^{|x|}_{-|x|}\,dx_n\Big)\,dx$$

Thus, thee integral you are after becomes $$n\int^\infty_{-\infty}e^{-|x|}2^{n-1}|x|^{n-1}\,dx=n2^n\int ^\infty_0x^{n-1}e^{-x}\,dx=n2^n\Gamma(n)=2^n n!$$

For another more geometric approach, similar to polar coordinates, notice that $\max_{1\leq j\leq n}|x_j|=\|x\|_\infty$ is a norm in $\mathbb{R}^n$. It is not difficult to check (via monotone class arguments) that for any norm $\rho$ on $\mathbb{R}^n$ and any nonnegative measurable function $f$ on $\mathbb{R}$ $$ \int_{\mathbb{R}^n}f(\rho(x))\,dx=n \lambda_n(\{x:\rho(x)\leq1\})\int^\infty_0 f(r) r^{n-1}\,dr$$ where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$.

For the particular norm $\|\,\|_\infty$, the volume of the $\|\;\|_\infty$-unit ball is $2^n$

I have another nice approach (hope you'll agree on that):

From symmetry on the sign:

$$\int_{\mathbb{R}^{n}}e^{-\max\{|x_1|,...,|x_n|\}}= 2^n\int_{\mathbb{R_{+}}^{n}}e^{-\max\{x_1,...,x_n\}}$$

By taking a permutation $0\leq x_1\leq x_2\leq...\leq x_n$, and taking the limit of the growing set $\lim_{k\to\infty}E_k = [0,k]^n$ we get

$$2^n\cdot n! \cdot \int_{\mathbb{R_{+}}^{n}\cap\{x\in\mathbb{R}^n|0\leq x_1\leq...\leq x_n\}}e^{-\max\{x_1,...,x_n\}} \\= 2^nn!\int_{\{0\leq x_1\leq...\leq x_n\}}e^{-x_n} \\= 2^nn!\lim_{k\to\infty} \int_0^k \int_{x_1}^k...\int_{x_{n-1}}^k e^{-x_n}$$

Im going to put $k=\infty$ just to make everything shorter. we get:

$$2^n\cdot n!\int_0^\infty \int_{x_1}^\infty...\int_{x_{n-1}}^\infty e^{-x_n} = 2^n\cdot n! \int_0^\infty e^{-x_1} = 2^nn! $$