Show $\hat\beta_1 = \frac{\sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^{n}(X_i - \bar{X})^2}$

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Given $$\hat\beta_1 = \frac{\sum_{i=1}^{n} X_i Y_i - \frac{1}{n} \sum_{i=1}^{n} X_i \sum_{i=1}^{n} Y_i}{\sum_{i=1}^{n} X_i^2 - \frac{1}{n} (\sum_{i=1}^{n} X_i^2)}$$

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$$\hat\beta_1 = \frac{\sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^{n}(X_i - \bar{X})^2}$$


Attempt:

$$ = \frac{\sum_{i=1}^{n} X_i Y_i - \frac{1}{n} \sum_{i=1}^{n} X_i \sum_{i=1}^{n} Y_i}{\sum_{i=1}^{n} X_i^2 - \frac{1}{n} (\sum_{i=1}^{n} X_i^2)} = \frac{\sum_{i=1}^{n} X_i (Y_i - \frac{1}{n} \sum_{i=1}^{n} Y_i)}{\sum_{i=1}^{n} X_i (\sum_{i=1}^{n} X_i - \frac{1}{n} \sum_{i=1}^{n} X_i)}$$

$$=\frac{\sum_{i=1}^{n} X_i (Y_i - \bar{Y})}{\sum_{i=1}^{n} X_i (\sum_{i=1}^{n} X_i - \bar{X})}$$

I'm confused after this.

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$$\sum_{i=1}^{n} X_i Y_i - \frac{1}{n} \sum_{i=1}^{n} X_i \sum_{i=1}^{n} Y_i = n\sum_{i=1}^n \frac{X_iY_i}{n} - n\sum_{i=1}^n\frac{X_i}{n}\sum_{i=1}^n\frac{Y_i}{n}=n(\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]) = nCov(X,Y)$$ $$\sum_{i=1}^nX_i^2-\frac{1}{n}(\sum_{i=1}^nX_i)^2=n\sum_{i=1}^n\frac{X_i^2}{n}-n\left( \sum_{i=1}^n\frac{X_i}{n}\right)^2=n(\mathbb{E}[X^2]-\mathbb{E}^2[X])=nVar(X)$$
And with this you got:
$$\hat\beta_1=\frac{nCov(X,Y)}{nVar(X)}=\frac{Cov(X,Y)}{Var(X)}=\frac{\sum_{i=1}^{n} (X_i - \bar{X})(Y_i - \bar{Y})}{\sum_{i=1}^{n}(X_i - \bar{X})^2}$$

So you got wrong written the denominator