# Why Test Statistic for the Pearson Correlation Coefficient is $\frac {r\sqrt{n-2}}{\sqrt{1-r^2}}$

I am learning hypothesis testing for Pearson Correlation Coefficient. The source did not explain why the test statistic $$\frac {r\sqrt{n-2}}{\sqrt{1-r^2}}$$ satisfy T distribution with $$n-2$$ degree of freedom.

Could anyone show me the assumption and proof?

When the residuals in a linear regression are normally distributed, the least squares parameter $$\hat{\beta}$$ is normally distributed. Of course, when the variance of the residuals must be estimated from the sample, the exact distribution of $$\hat{\beta}$$ under the null hypothesis is $$t$$ with $$n-p$$ degrees of freedom ($$p$$ the dimension of the model, usually two for a slope and intercept).
Per @Dason's link, the $$t$$ for the Pearson Correlation Coefficient can be shown to be mathematically equivalent to the $$t$$ test statistic for the least squares regression parameter by:
$$t = \frac{\hat{\beta}}{\sqrt{\frac{\text{MSE}}{\sum (X_i - \bar{X})^2}}}= \frac{r (S_y / S_x)}{\sqrt{\frac{(n-1)(1-r^2)S_y^2}{(n-2)(n-1)S_x^2}}}=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$$