Is $\hat\sigma^2=\frac{\sum e_i^2}{n-1}$ in simple linear regression with no interception

Question

I have this question. I know that when we have both parameters $\beta_1, \beta_0$, the estimator for the variance of the error is $\hat\sigma^2=\frac{\sum e_i^2}{n-2}$. I tried to prove that when we remove the intercept parameter, $\hat\sigma^2=\frac{\sum e_i^2}{n-1}$. But I lead that $\mathbb{E}(\hat\sigma^2)=\frac{1}{n-1}\left[\frac{\sum x_i^2}{\sum (x_i-\bar x)^2}\sigma^2+n\sigma^2-2\sigma^2\right]$. Any idea on how to proceed?