The Least Squares Estimators are Unbiased

From the previous posting, the least squares estimators
 1) Estimate of of $\beta_{1}$ : $b_{1}=\frac{\sum(X_{i}-\bar{X})(Y_{i}-\bar{Y})} {\sum(X_{i}-\bar{X})^2} = \frac{\sum(X_{i}-\bar{X})Y_{i}}{\sum (X_{i}-\bar{X})^2}=\frac{\sum X_{i}Y_{i}-n\bar{X}\bar{Y}}{\sum X_{i}^2-n\bar{X}^2}$

 2) Estimate of of $\beta_{0}$: $b_{0}=\bar{Y}-b_{1}\bar{X}$
 
They are unbiased!! why? $\Rightarrow$ Need to show that $E[b_{1}]=\beta_{1}$ and $E[b_{0}]=\beta_{0}$
Proof
First of all, there are three things we're going to use.
$k_{i}= \frac{X_{i}-\bar{X}}{\sum(X_{i}-\bar{X})^2}$, where $\sum k_{i}= 0, \ \sum k_{i}X_{i}=0, \ \sum k_{i}^2=\frac{1}{\sum(X_{i}-\bar{X})^2}$    
(1) $\sum k_{i}=0 \Rightarrow \frac{(X_{i}-\bar{X})}{S_{XX}}= \frac{\sum X_{i}-n\bar{X}}{S_{XX}}= \frac{n\bar{X}-n\bar{X}}{S_{XX}}=0$
(2) $\sum k_{i}X_{i}=1 \Rightarrow \frac{\sum( X_{i}-\bar{X})(X_{i}-\bar{X})}{S_{XX}}=\frac{\sum(X_{i}-\bar{X})^2}{S_{XX}}= \frac{S_{XX}}{S_{XX}}=1$
(3) $\sum k_{i}^2= \frac{1}{\sum(X_{i}-\bar{X})^2} \Rightarrow \sum k_{i}^2= \sum( \frac{X_{i}-\bar{X}}{S_{XX}})^2=\frac{1}{S_{XX}}\sum (X_{i}-\bar{X})^2=\frac{S_{XX}}{(S_{XX})^2}=\frac{1}{S_{XX}}$ 
 
Now, by using these results, we can finally show that the least squares estimators are unbiased!!
Proof
1) $E[b_{1}]= \beta_{1}$  
   $\Rightarrow$ $E[b_{1}]=E[\sum k_{i}Y_{i}]= \sum k_{i}E[Y_{i}]$ as $\sum k_{i}$ is a constant! And $Y_{i}=\beta_{0}+\beta_{1}X_{i}+\varepsilon _{i}$
                   $=\sum k_{i}(\beta_{0}+\beta_{1}X_{i})=\beta_{0} \sum k_{i}+ \beta_{1} \sum k_{i}X_{i}=\beta_{1}$

2) $E[b_{0}]=\beta_{0}$
  $\Rightarrow E[b_{0}]=E[\bar{Y}-b_{1}\bar{X}]=E[\bar{Y}]-E[b_{1}\bar{X}]=\beta_{0}+\beta_{1}\bar{X}-\bar{X}\beta_{1}=\beta_{0}$
        Notice that $Y_{i}$ is a random value, therefore $\bar{Y}$, $\sum k_{i}Y_{i}$ are also random!!
        However, $\bar{X}$ is NOT a random value, but it's a constant!
        *E(aY)=a E(Y), where a is a constant!