How to prove test statistic has $\chi^2$ distribution using minimum chi-square estimator? This is for iid case where the test statistic is 
$$\sum \frac{(\mathsf{observed} - \mathsf{expected})^2}{\mathsf{expected}}$$
It is supposedly shown on page 5 of Fisher (1924; pdf), but I do not understand his writing. I've tried Googling the proof as well. 
 A: Below you find the main line of reasoning of the proof in the paper you refer to:
In the paper $\chi^2$ is the test statistic $\chi^2= \sum_i \frac{(O_i-E_i)^2}{E_i}$ where $O_i$ are the observed cell counts and $E_i$ are the expected cell counts, assuming that the latter are ''truely'' known. 
The $\chi^{'2}$ is similar but slightly different; $\chi^{'2}= \sum \frac{(O_i-E'_i)^2}{E'_i}$, i.e. the expected counts are estimated and the estimates are $E'_i$. Estimates are used when the ''true'' $E_i$ are unknown, but these estimates introduce a ''source of error'' and Fisher analyses the consequences of that. 
The expected values are estimated via an effcient estimation of an underlying parameter whose ''true'' (but unknown - else we don't have to estimate it) value is $\theta$ and the estimated value is $\theta'$. The deviation between the true and the estimated value (estimation error) is $\delta \theta$. 
On page 448 of the paper you refer to Fisher shows that $\chi^{2}$ - $\chi^{'2}=\frac{(\delta \theta)^2}{\sigma^2}$ .
The right hand side of that equation is equal to $\left( \frac{\delta \theta}{ \sigma} \right)^2$ and the quantity between the brackets is equal to a normal variable divided by its standard deviation, so it is a squared standard normal variable or a chi-squared with one df. 
In other words, the difference between the test statistic using the ''true'' expected cell counts and the one using estimated expected cell counts is a chi-squared with one degree of freedom so we find that $\chi^{2}$ - $\chi^{'2}=\chi^2_1$, then, taking the expected value, and using the fact that the expected value of a chi-square is equal to its degrees of freedom, we find that $\nu - \nu' = 1$, where $\nu$ are the degrees of freedom of $\chi^{2}$ (i.e. using the true expected counts) and $\nu'$ are the degrees of freedom of $\chi^{'2}$ (i.e. using the estimated expected counts). 
So Fisher found that $\nu' = \nu - 1$, or, in words, if you replace the true expected cell counts by estimated expected cell counts, then you lose one degree of freedom for the chi-square. 
The result $\chi^{2}$ - $\chi^{'2}=\chi^2_1$ also learns that $\chi^{'2}$ is $\chi^{2} - \chi^2_1$ so a sum of squared standard normals, so it is also chi-square (see page 448). 
In the remainder of the paper he talks about the impact of using inefficient statistics and about the large sample assumption. 
Note also that Fisher refers to Elderton's table for finding critical values of chi-square; he did not have calculators as we do, so he had to use tables. 
Note also that, for contingency tables, for e.g. a $2 \times 2$ table, you estimate a $\theta_1$ for the rows and a $\theta_2$ for the columns so you lose a degree for the rows and one for the columns. 
