Stochastic Differential of $e^{B(t)-\frac{1}{2}t}$

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Stochastic processes beginner here!

So I'm trying to find the stochastic differential of $e^{B_t-\frac{1}{2}t}$

Ito's formula applied to $f[t, B_t]$, a function of time and Brownian motion, yields:

$$df=\frac{df}{dt}dt+\frac{df}{dB_t}dB_t+\frac{1}{2}\frac{d^2f}{dt^2}(dt)^2+\frac{d^2f}{dtdB_t}dtdB_t+\frac{1}{2}\frac{d^2f}{dB_t^2}(dB_t)^2$$

where, $$(dt)^2=0;dtdB_t=0;(dB_t)^2=dt,$$ and $$\begin{align} \frac{df}{dt}&=-\frac{1}{2}e^{B_t-\frac{1}{2}t}\\ \frac{df}{dB_t}&=e^{B_t-\frac{1}{2}t}\\ \frac{d^2f}{dB_t^2}&=e^{B_t-\frac{1}{2}t} \end{align}$$

Subbing into the equation: $$\begin{align} d\left(e^{B_t-\frac{1}{2}t}\right)&=-\frac{1}{2}e^{B_t-\frac{1}{2}t}dt+e^{B_t-\frac{1}{2}t}dB_t+\frac{1}{2}e^{B_t-\frac{1}{2}t}dt\\ &=e^{B_t-\frac{1}{2}t}dB_t \end{align}$$

Is my process and answer correct? Many thanks!