[theemathas]

The o(1) here represents a decreasing function in m, such as 1/m. (Note: This is little o, not big o.)

Therefore, for any k > 1: m^(1+o(1)) grows more slowly than m^k but faster than m

[im3w1l]

I'm trying to come up with an example to check my understanding. n (log n)^k for some constant k would qualify right?

[mauricioc]

Yes. (log n)/(n^epsilon) tends to zero for any positive epsilon, and so log n = n^(o(1)). The same holds for (log n)^k.