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Let $M_t={\max }_{0\le s\le t}{B}_s$. The Brownian reflection principle givesHere 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 thereforeSince $\Phi (h)-\Phi (0)\sim h/\sqrt{2\pi }$ as $h\downarrow 0$, the Brownian barrier survival asymptotic isAny 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.