I'm struggling with asymptotic notation a little bit...
As $k$ becomes large, are these two inequalities actually the same?
$k \le n^{\log (n+1) - 1}(1+o(1))$ and $k \le n^{\log (n)}(1+o(1))$?
the first one is better, but it makes no difference since log(n) is asymptotic to log(n+1)-1?
The expression on the LHS can be rewritten as $$ n^{\log (n+1) -1} = n^{\log n -1 + \log (1+\frac{1}{n})} \sim n^{\log n -1} n^{\frac{1}{n}} $$ The second term converges to 1 as $n \to \infty$, but the ratio $\frac{n^{\log n} }{n^{\log n -1}} = n^{\log n -\log n +1} = n$ which diverges to infinity.