Summation Proof I'm getting stuck halfway through this:
Show that $$\sum_{i=1}^n (y_i - \bar y_s)^2 = \sum_{i=1}^n (y_i)^2 - n\bar y_s^2$$
My skills with manipulating sums are quite rusty. I multiply the left side and distribute the sum to each part. I can see that the middle term needs to become $-2n\bar y_s^2$ in order to combine with the third term to make $-n\bar y_s^2$, but I can't quite get there. 

Suppose that 
$$\bar y_s=\frac 1n\sum_{i=1}^{n}y_i.$$
Then, we have$$\begin{align}\sum_{i=1}^{n}(y_i-\bar y_s)^2&=\sum_{i=1}^{n}(y_i^2-2\bar y_sy_i+\bar y_s^2)\\&=\sum_{i=1}^{n}y_i^2-\sum_{i=1}^{n}2\bar y_sy_i+\sum_{i=1}^{n}\bar y_s^2\\&=\sum_{i=1}^{n}y_i^2-2\bar y_s\sum_{i=1}^{n}y_i+\bar y_s^2\sum_{i=1}^{n}1\\&=\sum_{i=1}^{n}y_i^2-2\bar y_s\cdot n\bar y_s+\bar y_s^2\cdot n\\&=\sum_{i=1}^{n}y_i^2-n\bar y_s^2.\end{align}$$
