In class a while ago we used the following simplification:
$$ \mathbb E \left[ \exp\left(\langle \boldsymbol a,\mathbf W_t\rangle \right) \right] \quad =\quad \exp\left(\frac12 |\boldsymbol a|^2 t\right) $$
with $\boldsymbol a$ a constant $n$-dim vector, $\mathbf W_t$ an $n$-dim Brownian Motion.
I can recognize the quadratic variation on the right hand side and came up with the following (a bit hand wavy):
$$ \exp\left(\langle \boldsymbol a,\mathbf W_t\rangle \right) = \exp\left(\int_0^t d(\langle \boldsymbol a,\mathbf W_s\rangle) \right) \stackrel{\text{ito}}{=} \exp\left(\int_0^t \langle \boldsymbol a,d\mathbf W_s\rangle \right)\exp\left(\frac12 |\boldsymbol a|^2 t\right) $$
and then we'd need the expected value of the first term to be equal to $1$. But I don't see why this holds. It also looks fairly similar to the mgf of a normal distribution but again I don't see the connection clearly.
Let $W_t = (W_t^1,\ldots,W_t^n)$ a $n$-dimensional Brownian motion. Then the processes $(W_t^j)_{t \geq 0}$ are independent 1-dimensional Brownian motions ($j=1,\ldots,n$). Thus
$$\mathbb{E}\big(e^{\langle a,W_t \rangle} \big) = \prod_{j=1}^n \mathbb{E}\big(e^{a_j \cdot W_t^j} \big) \stackrel{W_t^j \sim N(0,t)}{=} \prod_{j=1}^n e^{\frac{1}{2} a_j^2 \cdot t} = e^{\frac{1}{2} |a|^2 \cdot t}$$
where we used $\mathbb{E}(e^{a \cdot X}) = e^{\frac{1}{2}a^2}$ for $X \sim N(0,1)$.