$S_{xx}$ and $S_{xy}$ in linear regression I am trying to prove that an equivalent way to perform the test for significance of regression
in multiple linear regression is to base the test on $R^2$ as follows:

to test $H_0:\beta_1 = \beta_2 = ... = \beta_k$ versus $H_1$ : at least one $\beta_j \neq 0$, calculate $$F_0=\frac{R^2(n-p)}{k(1-R^2)}$$ and to reject $H_0$ if the computed value of $F_0$ exceeds $F_{\alpha,k,n-p}$, where $p = k + 1$.

Now, I am starting from the easier case of simple linear regression, and so I want to show that $$t^2=\frac{R^2(n-2)}{1-R^2}.$$ I did the following.
$$t=\frac{\hat{\beta_1}}{se(\hat{\beta_1})}=\frac{S_{xy}}{S_{xx}}\sqrt{\frac{S_{xx}(n-2)}{SS_{res}}} \Rightarrow t^2= \frac{S^2_{xy}(n-2)}{S_{xx}SS_{res}}$$
and
$$R^2=\frac{SS_{reg}}{SS_{tot}}=\frac{S^2_{xy}}{S_{xx}SS_{tot}} \Rightarrow 1-R^2=\frac{S_{xx}SS_{tot} - S^2_{xy}}{S_{xx}SS_{tot}} \Rightarrow \frac{R^2(n-2)}{1-R^2}=\frac{S^2_{xy}(n-2)}{S_{xx}SS_{tot} - S^2_{xy}}.$$
How do I conclude? Thanks!
 A: Notation:
$$
S_{xy}=\sum_{i=1}^{n}(x_i-\bar x)(y_i-\bar y)
$$
and correspondingly for $S_{xx}$. $SS_{tot}=\sum_{i=1}^{n}(y_{i}-\bar{y})^{2}$, $SS_{reg}=\sum_{i=1}^{n}(\widehat{y}_{i}-\bar{y})^{2}$ and $SS_{res}=\sum_{i=1}^{n}\hat{u}_{i}^{2}$ stand for total (i.e., variation of the $y_i$), regression (i.e., variation of the fitted values) and residual sum of squares.
We have that
$$
R^2=\frac{S_{xy}^2}{S_{xx}SS_{tot}}
$$
and, from $SS_{tot}=SS_{reg}+SS_{res}$,
$$
R^2=1-\frac{SS_{res}}{SS_{tot}}
$$
so that
$$
SS_{tot}(1-R^2)=SS_{res}^2
$$
and hence
$$
t^2= \frac{S^2_{xy}(n-2)}{S_{xx}SS_{tot}(1-R^2)}.
$$
One could also have continued from your denominator $S_{xx}SS_{tot} - S^2_{xy}$ to get
$$
S_{xx}(SS_{reg}+SS_{res}) - S^2_{xy},
$$
so that it would remain to show
$$S_{xx}SS_{reg} - S^2_{xy}=0.$$
This follows from
$$
SS_{reg}=\frac{S^2_{xy}}{S_{xx}},
$$
which, in turn, follows from
$$\widehat{y}_{i}=a+bx_{i}$$ and
$$\widehat{y}_{i}-\bar{y}=b(x_{i}-\bar{x})$$ so that
$$SS_{reg}=b^{2}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}=b^{2}S_{xx}.$$
Plugging in $$b=S_{xy}/S_{xx}$$ gives
$$SS_{reg}=\frac{S^2_{xy}}{S_{xx}}$$
