We try to give meaning to the integral $\int_0^t W_s dW_s$. It is clear that this integral can not exists in Riemann-Stiltjes sense, because of the unbounded variation of $W_t$. Thus for each separate trajectory the integral sums \begin{equation} S_n = \sum_{i=1}^n W_{t_{t-1}} \Delta W_i = \sum_{i=1}^n W_{t_{t-1}} (W_{t_i} -W_{t_{i-1}}) \end{equation} By using \begin{equation*} (W_{t_i} - W_{t_{i-1}})^2 = W_{t_i}^2 + W_{t_{i-1}}^2 - 2 W_{t_i}W_{t_{i-1}} \end{equation*} It is possible to represent $S_n$ in the form \begin{equation} S_n = \frac{1}{2} W_t^2 - \frac{1}{2}Q_n(t). \end{equation} The mean-square convergence $S_n \rightarrow \frac{1}{2}W_t^2 - \frac{1}{2}t$ takes place. The limit $\frac{1}{2}W_t^2 - \frac{1}{2}t =: I_t$ is called stochastic integral and we write \begin{equation*} \int_0^t W_s dW_s = \frac{1}{2}W_t^2 - \frac{1}{2}t. \end{equation*}
Now my problem is behind the algebra, I can't get from (1) to (2).
This is where I am so far \begin{align*} \sum_{i=1}^n W_{t_{i-1}} (W_{t_i} -W_{t_{i-1}}) &= \sum_{i=1}^n W_{t_{i-1}}W_{t_i} - W_{t_{t-1}}^2 =\\ &= \sum_{i=1}^n -\frac{1}{2}(W_{t_i} -W_{t_{i-1}})^2 + \frac{1}{2}W_{t_i}^2 + \frac{1}{2}W_{t_{i-1}} - W_{t_{i-1}}^2 =\\ &= -\frac{1}{2} \sum_{i=1}^n (W_{t_i} -W_{t_{i-1}}) + \frac{1}{2} \sum_{i=1}^n W_{t_i}^2 - W_{t_{i-1}}^2\\ &= - \frac{1}{2}Q_n(t) + \frac{1}{2} \sum_{i=1}^n W_{t_i}^2 - W_{t_{i-1}}^2 \end{align*}
You are missing the very last step which is to note that \begin{equation} \sum\limits_{i=1}^n(W_{t_i}^2 - W_{t_{i-1}}^2) = (W_{t_n}^2 - W_{t_{n-1}}^2) + (W_{t_{n-1}}^2 - W_{t_{n-2}}^2) + ... + (W_{t_1}^2 - W_{t_0}^2) = W_{t_n}^2 - W_{t_0}^2 = W_t^2 \end{equation} because $t_n = t$, $t_0 = 0$ and $W_0 = 0$.