= Solution
Let $M_t=\max_{0\leq s\leq t}B_s$. The <Brownian reflection principle> gives
$$
\mathbb P(M_t>1)=2\mathbb P(B_t>1).
$$
Here is the reflection argument: on paths that hit $1$ before $t$, reflect all increments after their first hitting time. The <Strong Markov property> gives <independent> <Brownian motion> after that time, whose sign can be reversed without changing its law. This exchanges paths ending below $1$ with paths ending above $1$. The latter have necessarily hit $1$, and the endpoint has no atom at $1$, giving the displayed equality. It also shows that $M_t$ has a continuous distribution at every positive level.
Writing $\Phi$ for the <standard normal distribution function>, the <probability> of staying below the barrier is therefore
$$
\mathbb P(B_s\leq1\text{ for all }s\leq t)=2\Phi(t^{-1/2})-1.
$$
Since $\Phi(h)-\Phi(0)\sim h/\sqrt{2\pi}$ as $h\downarrow0$, the <Brownian barrier survival asymptotic> is
$$
\boxed{\alpha=\tfrac12,\qquad \lim_{t\to\infty}\sqrt t\,\mathbb P(B_s\leq1\text{ for all }s\leq t)=\sqrt{\frac2\pi}.}
$$
Any smaller exponent gives a zero limit and any larger exponent gives an infinite limit, so this is the unique exponent producing a finite positive constant.
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