If $Z_t$ is a progressive process and $X_t$ is defined by $X_t = 1+\int_0^t Z_s dB_s$, then use Ito's formula to $Y_t=ln(X_t)$, to show that $$ X_t = \exp\Bigl(\int_0^t \frac{Z_s}{X_s} dB_s - \frac{1}{2}\int_0^t \bigl(\frac{Z_s}{X_s}\bigr)^2 ds \Bigr)$$
Consider $Y_t = ln(X_t) = \int_0^t \frac{Z_s}{X_s} dB_s - \frac{1}{2}\int_0^t \bigl(\frac{Z_s}{X_s}\bigr)^2 ds $.
So $$dY_t = \frac{Z_t}{X_t} dB_s - \frac{1}{2}\bigl(\frac{Z_t}{X_t}\bigr)^2 $$ and $d\langle Y \rangle_t = (\frac{Z_S}{X_s})^2. $ Then by Ito's formula:
$$ X= df(Y_t) = f'dY_t+\frac{1}{2}f''d\langle Y\rangle_t$$ $$ = \frac{1}{Y_t}dY_t - \frac{1}{2Y_t^2}\left(\frac{Z_S}{X_s}\right)^2dB_s$$
This is clearly completely wrong, how should I have approached this instead...? My end goal is to get this in the form $X_t = 1+\int_0^t Z_s dB_s$.
From Ito's formula applied to $Y_t := \ln(X_t)$, we have
\begin{align*} dY_t &= \frac{1}{X_t}dX_t - \frac 12 \frac{1}{X_t^2}d\langle X,X\rangle_t \\ &= \frac{Z_t}{X_t} dB_t - \frac 12 \left( \frac{Z_t}{X_t} \right)^2dt \end{align*}
so $\ln(X_t) = \ln(X_0) + \int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t(\frac{Z_s}{X_s})^2ds = \int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t(\frac{Z_s}{X_s})^2ds.$ Now just exponentiate both sides: $$X_t = \exp\left(\int_0^t \frac{Z_s}{X_s}dB_s - \frac 12 \int_0^t\left(\frac{Z_s}{X_s}\right)^2ds\right)$$