How can I arrive from $\prod_{i=1}^{D} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i$ to $\pi^{D/2}$? This is part of the excersise 1.18 of the Bishop's Book, Pattern Recognition and Machine Learning. I attached the excersise and the information. I do not know how he using the result (1.126) in  the left side of (1.142) arrives to $\pi^{D/2}$ (That is  the solution he provides). I tried several ways but I get nothing.
1.126: $I=(2 \pi \sigma^{2})^{1/2}$
which comes from:
$I^{2}= \int_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^{2}}x^{2}-\frac{1}{2\sigma^{2}}y^{2}}dxdy $
And solving transforming form Cartesian coordinates to polar ones, substituing u = $r^{2}$ and performing the integrals over u and $\theta$

I attached the Excersises, 1.18. I can not upload more pictures so here is a drive link to the others: https://drive.google.com/drive/folders/1sc6WFyhb5wEevZOv-6qcTFKuoieCxPGH?usp=sharing
\begin{align*}
I &amp;=\int_{-\infty}^{\infty} e^{-x_i^2} dx_i \\
I^2 &amp;= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x_i^2}  e^{-y_i^2} dx_i dy_i \\
 &amp;=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}  r dr d\theta \tag{1} \\
 &amp;=\int_{0}^{2\pi} -\frac{e^{-r^2}}{2} \bigg \rvert_{r=0}^{\infty} d\theta \\
 &amp;=\int_{0}^{2\pi} \frac{1}{2}  d\theta \\
I &amp;=\sqrt{\pi} \\
\end{align*}
And so,
\begin{align*}
\prod_{i=1}^D \int_{-\infty}^{\infty} e^{-x_i^2} dx_i  &amp;= \prod_{i=1}^D \sqrt{\pi} \\
&amp;= \boxed{\pi^{\frac{D}{2}}}
\end{align*}
Where in $(1)$ the double integral was converted to polar coordinates.
Recall that $\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi} = \pi^{1/2}.$ Therefore
$$
\prod_{i=1}^{D} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i
= \prod_{i=1}^{D} \pi^{1/2}
= (\pi^{1/2})^D
= \pi^{D/2}.
$$