test statistic for hypothesis testing for population proportion with population proportion unknown

Question

In hypothesis testing for population mean, if population variance is known, then sampling distribution of sample mean $$\frac{\bar{X}-\mu } {\frac{\sigma}{\sqrt n}} \sim Z\left ( 0,1^2 \right )$$ If population variance is unknown, then $$\frac{\bar{X}-\mu } {\frac{s_{n-1}}{\sqrt n}} \sim t_{n-1}$$ There is a rigorous proof for this. Basically if there are a standard normal random variable $Z$ and a chi-squared random variable $\chi_n^2$. Then $$\frac{Z}{\sqrt{\frac{\chi_n^2}{n}}}\sim t_n$$ and $$\frac{\bar{X}-\mu } {\frac{s_{n-1}}{\sqrt n}} =\frac{\frac {\bar{X}-\mu } {\frac {\sigma} {\sqrt n}}} {\sqrt\frac{{\frac{\left ( n-1 \right )s_{n-1}^2}{\sigma^2\ }}}{\left ( n-1 \right )}} =\frac{Z}{\sqrt{\frac{\chi_{n-1}^2}{n-1}}} \sim t_{n-1}$$
Similarly in the hypothesis testing for population proportion, if the population proportion is known, then the sampling distribution of sample proportion under CLT $$\frac{\hat{p}-p } {\sqrt{\frac{p(1-p)}{n}}} \sim Z\left ( 0,1^2 \right )$$ If population proportion is unknown, then why still $$\frac{\hat{p}-p } {\sqrt{\frac{\hat p(1-\hat p)}{n}}} \sim Z\left ( 0,1^2 \right )$$ Shouldn't it be a different distribution? Is there a rigorous proof for the above result?