This is essentially a "Laplace Method" argument that allows you to conclude that :
$\left(\int_0^te^{B_s}ds\right)^\frac{1}{\sqrt{t}}$ converges in law to $exp(S_1)=e^{\sup_{s\leq 1}\textit{B}_s}$.
You have the following equalities (sometime only in law):
$(\int_0^te^{B_s}ds)^\frac{1}{\sqrt{t}}= (t\int_{0}^{1}e^{\sqrt{t}\textit{B}_s}ds)^\frac{1}{\sqrt{t}}=t^\frac{1}{\sqrt{t}}.||e^{\textit{B}_.}||_{L^\sqrt{t}[0,1]} $
Laplace's Method allows you to conclude that $\lim_{t\to \infty}||e^{\textit{B}_.}||_{L^\sqrt{t}[0,1]}=||e^{\textit{B}_.}||_{L^\infty[0,1]}=e^{\sup_{s\leq 1}\textit{B}_s}$
The only thing you need to apply Laplace Method is that the function is $L^{\infty}$ which is true for a.s. for the paths of $e^{\textit{B}_.}$ over the interval $[0,1]$
Regards
Edit proof of Laplace Method : Let's take a positive $f\in L_\mu^{\infty}(X)$, then first we have for any $p>1$ :
$\|f\|_{L^p}\leq \|f\|_{L^{\infty}}\cdot (\mu(X))^{1/p}$ (here $\mu(X)=1$ but this works for finite measures aswell)
Which gives half of the result.
Second, let's fix $\epsilon>0$, and note $A_\epsilon=\{x \in X s.t.|f(x)|\geq \|f\|_{L^{\infty}}-\epsilon\}$, and such that $\mu(A_{\epsilon})>0$ (there exists such an $\epsilon$ and for every values inferior to this value, the corresponding set has non zero measure, this can be seen by absurd)
Then we have:
$\int_{X}{|f(s)|}^p d\mu(s)=\int_{A_{\epsilon}}{|f(s)|}^p d\mu(s)+\int_{A_{\epsilon}^c}{|f(s)|}^p d\mu(s) \geq (\|f\|_{L^{\infty}}-\epsilon)^p \cdot \mu(A_{\epsilon}) $
Taking the p-th root we get :
$(\|f\|_{L^{\infty}} - \epsilon)\cdot (\mu(A_{\epsilon}))^{1/p} \leq \|f\|_{L^p}$
Letting $p$ going to $\infty$ (note that $\epsilon$ is still fixed) we get :
$ \lim_{p \rightarrow +\infty}(\|f\|_{L^{\infty}} - \epsilon)\cdot (\mu(A_{\epsilon}))^{1/p} = \|f\|_{L^{\infty}} - \epsilon \leq \lim_{p \rightarrow +\infty} \|f\|_{L^p} (*)$
As this is true for any $\epsilon>0$ we get the desired result.
$(*)1\ge\mu(A_{\epsilon})>0$