I need to compute the differential $dI$ of $I(t) = e^{\sigma W(t)}$ where $W(t)$ is a brownian motion.
With $I(W_t) = e^{\sigma W_t}$ I can apply Itô's lemma and get $dI =\frac{\partial I}{\partial W_t}dW_t +\frac{1}{2} \frac{\partial^2I}{\partial W^2_t}dt=\sigma e^{\sigma W_t}dW_t+\frac{1}{2}\sigma ^2e^{\sigma W_t}dt$.
With $I(t)$ how can I get $dI$?
This is just an application of Ito's calculus.
Let $dX_t=\tilde \mu(X_t)dt+\tilde\sigma(X_t)dW_t$ for some approprtiate functions $\tilde\mu,\tilde\sigma$. From Ito's formula we get that for some $C^2$ function $f$ we have that
$$df(X_t) = \dfrac{\partial f}{\partial x}dX_t+\dfrac{1}{2}\dfrac{\partial^2 f}{\partial x^2}\tilde{\sigma}(X_t)^2dt.$$
Now let $f(X_t)=\exp(\sigma X_t)$ with $X_t=W_t$, then we have that $dX_t=dW_t$, and hence $\tilde\mu(\cdot)=0$ and $\tilde\sigma(\cdot)=1$. Therefore,
$$df(X_t) = \sigma \exp(\sigma X_t)dW_t+\dfrac{\sigma ^2}{2}\exp(\sigma X_t)dt.$$
If we let $I_t=f(X_t)$, we then obtain
$$dI_t = \frac{1}{2}\sigma ^2I_tdt + \sigma I_tdW_t,\ I_0=1$$
which is the equation of a geometric Brownian motion. You can verify that the formula is correct using the explicit formula for the solution of the geometric Brownian motion.