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No, you only have $|\alpha^{n_i-n_{i-1}} - 1|<A^{-n_{i-1}}$. Since it might be that the sequence of the $n_i$ is extremely sparse, this bound is a lot weaker than the required bound $A^{-(n_i-n_{i-1})}$.
For all $r$ in the range given we have $\frac{rn}{2\log_2 n}>\frac{r^{\log_2^c r}}{2\log_2 r}$, so almost all rank $r$-matrices with entries in $\{0,1\}$ satisfy $1'_nA1_n >\frac{rn}{2\log_2 n}>\frac{r^{\log_2^c r}}{2\log_2 r}$.
@Stefan Kohl: Ivic told me that the Barsky-Benzaghou-proof is philosophically correct in the sense that although it does not prove the conjecture, it gives a reason why the conjecture should be true. I haven't looked at the paper itself, though.
