I have to show that $$\mathbb{E}\left[ \int_0^t a B_s^{a-1} \ dB_s \right] = 0$$ I started with the fact that, if I show that the inner integral $\int_0^t a B_s^{a-1} \ dB_s$ is a martingale I get the wanted result.
One way to show that the stochastic integral is a martingale is to show that: $$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] < \infty$$ which is basically saying $aB_s^{a-1} \in \Lambda^2(t)$ (to have this we also need the process to be progressive, but it follows by BM properties I guess).
Using Fubini I can interchange expectation and integral, bring outside constants and write: $$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] = \int_0^t a^2 \ \ \mathbb{E}\left[ B_s^{\ 2(a-1)} \right] ds< \infty$$ any hint to show this? I should not use results on Brownian motion moments.
NOTE: edited after @surb comment!!
Here is a proof for $\int_0^{t} aEB_s^{a-1}ds <\infty$ which you wanted . You need to modify this based on the comment of Surb.
$B_s \sim \sqrt s X$ where $X$ has standard normal distribution. Hence, your integral $\int_0^{t} aEB_s^{a-1}ds <\infty$ is a constant times $\int_0^{t} s^{(a-1)/2}ds$ and this last integral equals $\frac {t^{(a+1)/2}} {(a+1)/2}$ if $a>-1$.