I cannot think of an easy way to push the OP's approach to completion. Here's a fresh approach.
Since $ \left( 1 + \frac{1}{n} \right)^c - 1 = O\left(\frac{1}{n} \right) \tag{1} $ for any fixed $c \in (0, 1)$, it follows that $ (n+1)^c - n^c = O(n^{c-1}) = o(1). $
Actually, $(1)$ needs some justification, which might be suppressed depending on the context (and the level of the author/intended audience).
A solution without big-Oh. As per the OP's comment below, here's a way to remove the use of the big-Oh notation in the proof. @Did's answer gives a similar proof that in fact gives a tighter estimate.
If $0< c < 1$, then we have: $ (n+1)^c-n^c = n^c \left[ \left( 1 + \frac{1}{n} \right)^c - 1 \right] \leq n^c \left[\left( 1 + \frac{1}{n} \right)^{1} - 1 \right] = n^{c-1}. $ We can proceed exactly as before.
Using Taylor expansion. This solution is from the OP's comment below. For small $x$, by Taylor expansion, we have $(1+x)^c = 1+cx + o(x)$. Therefore, $ (n+1)^c-n^c = n^c \left[ \left( 1 + \frac{1}{n} \right)^c - 1 \right] = n^c \left( \frac{c}{n} + o\left( \frac 1n \right) \right) = c n^{c-1} (1+o(1)).$