In the context of multivariate normal variable:
Let $X = (X_1 , \ldots , X_n )^T$ denote a vector of $n$ independent, standard normal random variables. Let $\bar{x} = (x_1 , \ldots , x_n )^T$ be a vector of real values. The joint cumulative distribution function for $X$ is given as follows, where $\bar{w} = (w_1 , \ldots , w_n )^T$
$$ \Pr(X \leq \bar{x}) = \prod_{i = 1}^{n} \left( \frac{1}{\sqrt{2 \pi}} \int_{w_i = - \infty}^{x_i} \exp \left( \frac{- w_i^2}{2} \right) \mathrm{d}w_i \right) $$
$$ = \left( \frac{1}{\sqrt{2 \pi}} \right)^n \int_{w_1 = - \infty}^{x_1} \ldots \int_{w_n = - \infty}^{x_n} \exp \left( \frac{- \bar{w}^T \bar{w}}{2} \right) \mathrm{d}w_1 \ldots \mathrm{d}w_n $$
My question: How did we obtain the last multiple integral?
Edit: Here's a rewritten version without the dot product:
$$ = \left( \frac{1}{\sqrt{2 \pi}} \right)^n \int_{w_1 = - \infty}^{x_1} \ldots \int_{w_n = - \infty}^{x_n} \exp \left( - \frac{1}{2} \left( w_1^2 + w_2^2 + \ldots + w_n^2 \right) \right) \mathrm{d}w_1 \ldots \mathrm{d}w_n $$
So how do we tackle this multiple integral to get the product above?
Summarizing the answer (many thanks to @Semiclassical ):
Using Fubini's theorem, we can prove that for a product of two functions $f(x)$, $g(y)$
$$ \iint \limits _{X\times Y}f(x) g(y)\,{\text{d}}x\, \text{d}y = \int_X f(x) \text{d}x \int_Y g(y) \text{d}y \\ $$
(Proof can be found for example here: https://math.stackexchange.com/a/699699/682779 )
So we can rewrite the CDF (applying the above theorem on line 3):
$$ \begin{aligned} & \Pr(X \leq \bar{x}) = \\ &= \prod_{i = 1}^{n} \left( \frac{1}{\sqrt{2 \pi}} \int_{w_i = - \infty}^{x_i} \exp \left( \frac{- w_i^2}{2} \right) \mathrm{d}w_i \right) \\ &= \left( \frac{1}{\sqrt{2 \pi}} \right)^n \prod_{i = 1}^{n} \left( \int_{w_i = - \infty}^{x_i} \exp \left( \frac{- w_i^2}{2} \right) \mathrm{d}w_i \right) \\ &= \left( \frac{1}{\sqrt{2 \pi}} \right)^n \int_{w_1 = - \infty}^{x_1} \ldots \int_{w_n = - \infty}^{x_n} \exp \left(\frac{- w_1^2}{2}\right) \exp \left(\frac{- w_2^2}{2}\right) \cdots \exp \left(\frac{- w_n^2}{2}\right) \mathrm{d}w_1 \ldots \mathrm{d}w_n \\ &= \left( \frac{1}{\sqrt{2 \pi}} \right)^n \int_{w_1 = - \infty}^{x_1} \ldots \int_{w_n = - \infty}^{x_n} \exp \left( - \frac{1}{2} \left( w_1^2 + w_2^2 + \ldots + w_n^2 \right) \right) \mathrm{d}w_1 \ldots \mathrm{d}w_n \\ &= \left( \frac{1}{\sqrt{2 \pi}} \right)^n \int_{w_1 = - \infty}^{x_1} \ldots \int_{w_n = - \infty}^{x_n} \exp \left( - \frac{1}{2} \bar{w}^T \bar{w} \right) \mathrm{d}w_1 \ldots \mathrm{d}w_n \end{aligned} $$ which is what we wanted.