Define Brownian motion as a continuous process $(B_t)$ with independent increments, such that $B_{s+t}-B_{s}$ has normal distribution with mean $0$ and variance $t$. How do you use the independence of increments to prove that for $t>0$ and $s_1<\cdots <s_n$ $$(B_{s_1t},\ldots,B_{s_nt})=_d(t^{1/2}B_{s_1},\ldots,t^{1/2}B_{s_n})$$ where $=_d$ means that both vectors have the same distribution.
Only using the independence of increments, not that the process is Gaussian.
(I'll only write this down for $n=2$.) Let $f$ be a bounded continuous function of two real variables; let $Z_1$ and $Z_2$ be independent standard normal random variables. Then $$ \eqalign{ \Bbb E[f(B_{s_1t},B_{s_2t})] &=\Bbb E[f(B_{s_1t},B_{s_1t}+(B_{s_2t}-B_{s_1t}))]\cr &=\Bbb E[f(\sqrt{s_1t}Z_1,\sqrt{s_1t}Z_1+(\sqrt{s_2t-s_1t}Z_2))]\cr &=\Bbb E[f(\sqrt{t}\sqrt{s_1}Z_1,\sqrt{t}\sqrt{s_1}Z_1+\sqrt{t}(\sqrt{s_2-s_1}Z_2))]\cr &=\Bbb E[f(\sqrt{t}B_{s_1},\sqrt{t}B_{s_1}+\sqrt{t}(B_{s_2}-B_{s_1}))]\cr &=\Bbb E[f(\sqrt{t}B_{s_1},\sqrt{t}B_{s_2})].\cr } $$ Of course it is crucial here that (because $B_s$ is normal with mean $0$ and variance $s$) $B_s$ has the same distibution as $\sqrt{s}Z$, where $Z$ is standard normal. Because $f$ was arbitrary, this shows that $(B_{s_1t},B_{s_2t})$ has the same distribution as $(\sqrt{t}B_{s_1},\sqrt{t}B_{s_2})$.