I ask and answer this question to provide an elementary proof for a known result for readers, as I personally could not find a good source with details that is reliably and freely available. You may add new proofs or connections with other properties.
A nice property about a random point $(X,Y)$ with independent and normally distributed Cartesian coordinates $X,Y \sim N(0,1)$ is that its polar coordinates: distance $R$ and angle $\Theta$, are also independent. More specifically, we have the following properties:
1- $R$ and $\Theta$ are independent.
2- $R$ has a Rayleigh distribution whose parameter is 1. Hence, $R^2$ has the chi-squared distribution with 2 degrees of freedom (which is the exponential distribution with mean 2).
3- $\Theta$ has the uniform distribution on the interval $[0,2\pi[$ (or $[-\pi,\pi[$, depending on how the angle is defined in the polar coordinate system).
Prove the above properties.
It is worth noting that the independence of the polar coordinates also follows from a characterization of normal distributions given below.
Let $(X,Y)$ be a random point where $X,Y$ are independent and have differentiable densities $f_X(x), f_Y(y)$. Then,
$f_{X,Y}(x,y)=f_X(x)f_Y(y)$ is radial, i.e., only depends on the distance between that point and the origin: $\sqrt{x^2+y^2}$, if and only if $X,Y \sim N(0,\sigma^2).$
One can see that
$$R=\sqrt{X^2+Y^2}$$ $$\Theta=2\pi-\text{atan2}\left(\frac{Y}{X} \right )$$
where atan2 is a modification of the arctangent function used for measuring the angle of any point on a circle. Note we use
$$\Theta=\text{atan2}\left( \frac{Y}{X} \right ) $$
if the angle is measured in the range $[-\pi,\pi]$.
One can write $X,Y$ in terms of $R, \Theta $ as
$$X = R\cos\Theta, Y= R\sin\Theta.$$
Hence, the joint density of $R, \Theta $ is given by
$$f_{R,\Theta}(r,\theta)= \big |\det \left (\text{Jacobian}_{r,\theta}(x,y) \right ) \big | f_{X,Y}(x, y) \quad \text{with} \, x=r\cos\theta, y=r\sin\theta$$
Considering $x=r\cos\theta, y=r\sin\theta$,
$$\text{Jacobian}_{r,\theta}(x,y)=\frac{\partial(x, y)}{\partial(r, \theta)} = \begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & \phantom{-}r\cos\theta \end{bmatrix},$$
so $$ \det \left (\text{Jacobian}_{r,\theta}(x,y) \right ) = r.$$
Moreover, we have
$$f_{X,Y}(x,y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}=\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}.$$
These together give
$$f_{R,\Theta}(r,\theta)= \frac{1}{2\pi}r e^{-\frac{r^2}{2}}, \quad r \ge 0, \, \theta \in [0,2\pi[. $$
One can see that this joint density is the product of the two densities of $\Theta \sim \text{Uniform}[0,2\pi[$ and $ R \sim \text{Rayleigh}(1)$, which completes the proof.