Refering from [1], given a set of $N$ data points $D = \{x_i\}$ drawn independently from a Gaussian with mean $\mu$ and standard deviation $\sigma$, the likelihood for these data is $$P(D|\mu, \sigma) = \prod_{i = 1}^{N}\dfrac{1}{\sigma \sqrt{2\pi}}\exp{\bigg[ -\dfrac{(x_i - \mu)^2}{2\sigma^2}\bigg]}.$$
How is this equation equivalent to
$$P(D|\mu, \sigma) = \dfrac{1}{\sigma^N {(2\pi)}^{N/2}}\exp{\bigg[ -\dfrac{1}{2\sigma ^2} \sum_{i=1}^{N}(x_i - \mu)^2\bigg]}?$$
Is there a simple proof that these two expressions are equivalent?
REFERENCE
[1] Bailer-Jones, C. (2017), Practical Bayesian Inference: A Primer for Physical Scientists, Cambridge: Cambridge University Press, p. 126.