This post discusses how to convert a daily standard deviation into an annualized one.
However, it doesn't include a proof.
I've tried to prove it myself by using a summation of Gaussians and trying to calculate the standard deviation of the set, but I'm getting stuck, and I'm sure there's an easier way.
The question- why is:
$$\sigma_{\rm annual} = \sigma_{\rm measured} \sqrt{T}$$
You don't need to assume each sample point is drawn from a Gaussian distribution, or even that they have the same distribution. It is enough for them to be pairwise uncorrelated and have the same standard deviation. If $\ D_1, D_2,\dots,D_T\ $ are pairwise uncorrelated with means $\ d_1,d_2,\dots, d_T\ $ and standard deviations $\ \sigma_1,\sigma_2,\dots,\sigma_T\ $, then the variance of $\ \sum_\limits{i=1}^TD_i\ $ is \begin{align} \sigma_\text{annual}^2&=\mathbb{E}\left(\left(\sum_{i=1}^T\big(D_i-d_i\big)\right)^2\right)\\ &=\mathbb{E}\left(\sum_{i=2}^T \sum_{j=1}^{i-1} \big(D_i-d_i\big)\big(D_j-d_j\big)\right)\\ &\hspace{2em}+\mathbb{E}\left(\sum_{i=1}^T \big(D_i-d_i\big)^2\right)\\ &\hspace{2em}+\mathbb{E}\left(\sum_{i=1}^{T-1} \sum_{j=i+1}^T \big(D_i-d_i\big)\big(D_j-d_j\big)\right)\\ &=\sum_{i=1}^T \mathbb{E}\left(\big(D_i-d_i\big)^2\right)\\ &=\sum_{i=1}^T \sigma_i^2\ , \end{align} the first and third terms on the right of the second equation's vanishing because $\ D_1, D_2,\dots,D_T\ $ are pairwise uncorrelated. Thus, if $\ \sigma_i=\sigma_0\ $ for all $\ i\ $, then \begin{align} \sigma_\text{annual}&=\sqrt{T \sigma_0^2}\\ &=\sigma_0\sqrt{T} \end{align}
