Let $X$ be a good integrator with $X_0=0$, then the process \begin{equation*} Z_t=\exp(X_t-\frac{1}{2}[X,X]_t)\prod_{0\leq s \leq t}(1+\Delta X_s) \exp(-\Delta X_s + \frac{1}{2}(\Delta X_s)^2) \end{equation*} satisfies $Z_t=1+(Z_{-} \circ X)$ and is called stochastic exponential. ($Z_{-} \circ X)$ is the stochastic integral of $Z_{-}$ with respect to X) To proof this I defined $X_t^1 = X_t - \frac{1}{2} [X,X]_t$ and $X_t^2 = \prod_{0\leq s \leq t} (1+\Delta X_s + \frac{1}{2}(\Delta X_s)^2)$. Then I have shown that $X_t^2$ is an adapted, cadlag process of finite variation and therefore a good integrator.
For the next step of the proof I wanted to use Ito's formula for the function $Z_t=\exp(X_t^1)X_t^2$
It gives \begin{align} Z_t = {} & 1 + (Z_{-} \circ X^1) + (\exp(X_{-}^1) \circ X^2) + \frac{1}{2}(Z_{-} \circ [X^1,X^1]) \\ & {} + \sum_{0 \leq s \leq t} (Z_s - Z_{s_{-}} - Z_{s_{-}} \, \Delta X_s^1 - \exp(X_{s_{-}}) \, \Delta X_s^2 - \frac{1}{2} Z_{s_{-}}(\Delta X^1)^2)\\[10pt] = {} & 1 + (Z_{-} \circ X) + (\exp(X_{-}^1) \circ X^2) + \sum_{0 \leq s \leq t} (Z_s - Z_{s_{-}} - Z_{s_{-}}\Delta X_s^1 - \exp(X_{s_{-}})\,\Delta X_s^2 - \frac{1}{2} Z_{s_{-}}(\Delta X^1)^2) \end{align} where I used in the equation that $(Z_{-}\circ X^1) = (Z_{-} \circ X) - \frac{1}{2} (Z_{-} \circ [X,X])$.
I know that $\Delta Z_s = (1+\Delta X_s)$. But how can I show that the previous equation $Z_t=\cdots= 1 + (Z_{-} \circ X)$?
The theory that I use is based on Philip E. Protter. Stochastic integration and differential equations. This example is also in his book, but I wanted to make a little different proof.
Thanks a lot for your efforts.