 Solution

On a one-dimensional representation, $s_i$ acts by a nonzero scalar ${\epsilon }_i$. Its square is one. The braid relation becomes ${\epsilon }_i^2{\epsilon }_{i+1}={\epsilon }_{i+1}^2{\epsilon }_i$, and hence ${\epsilon }_{i+1}={\epsilon }_i$. Thus all adjacent transpositions act by one common scalar $\epsilon$ with ${\epsilon }^2=1$.

In characteristic different from two, $\epsilon =1$ or $-1$. Since adjacent transpositions generate the group, these choices give exactly the trivial representation and sign representation. For $n\le 1$ they coincide, leaving only the trivial representation.