I have to verify, using only the definition of stochastic integral, that: $$ (B \cdot B)_t := \int_0^t B_s \text{d}B_s = \frac{1}{2} B_t^2 - \frac{1}{2}t. $$
Below what I've tried so far.
Given $M \in H^2$, $K \in L^2(M)$ the stochastic integral $K \cdot M$ is the only process such that: $$ \left< K \cdot M, N \right> = K \cdot \left< M, N \right> \quad \forall N \text{ martingale in } H^2. $$
Let $Y_t = \frac{1}{2} B_t^2 - \frac{1}{2}t$. By definition of stochastic integral I have to verify that $$ \left< B \cdot B, N \right> = B \cdot \left<B,N \right> \quad \forall N \in H^2 $$ namely $$ \left<Y,N \right> = B \cdot \left<B,N \right>. $$ Letting $Z_t = t$ it is equivalent to show that: $$ \left<\frac{1}{2} B^2,N \right> + \left<\frac{1}{2} Z,N \right> = B \cdot \left<B,N \right> $$ by bilinearity of covariation. Given the fact that $Z$ is a finite variation process (because it is continuous and increasing) I get $$ \left<\frac{1}{2} Z, N \right> = 0. $$ But now I'm stuck because I don't understand how to further manipulate $\left<B^2,N \right>$.
Any hint would be appreciated.
My favourite proof is this: \begin{align} &\int_0^tB_s\,dB_s=\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^nB_{t_i}(B_{t_{i+1}}-B_{t_i})\\ &=\frac{1}{2}\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^n(B_{t_i}-B_{t_{i+1}}+B_{t_{i+1}}+B_{t_i})(B_{t_{i+1}}-B_{t_i})\\ &=-\frac{1}{2}\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^n(B_{t_{i+1}}-B_{t_i})(B_{t_{i+1}}-B_{t_i})}_{\textstyle\langle B\rangle_t=t}+\frac{1}{2}\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^n(B^2_{t_{i+1}}-B^2_{t_i})}_{\textstyle B_t^2}\, \end{align} where $\Pi_n=\max\limits_{i=1,...,n}|t_{i+1}-t_i|$ and the last simplifcation to $B_t^2$ is due to the sum being telescoping.