Proving Convergence: Real s, 0<s<1, n \to \infty

[disregardthat]

How can we prove that $$n^s-(n-1)^s$$ converge to zero as $$n \to \infty$$ where s as a real number satisfies $$0<s<1$$?

I am specifically looking for a more or less elementary proof for this for real s. I think we can use the infinite binomial expansion, but I am looking for something that does not require more than elementary calculus.

 

[Gerenuk]

I thought about the graph of that function and why the limit is "obvious" from the graph. Translating the graph picture into mathematics, I think an easy way is the mean value theorem.
With it you find
$$n^s-(n-1)^s=s(n+\xi)^{s-1}\to 0$$

 

[disregardthat]

I thought about the graph of that function and why the limit is "obvious" from the graph. Translating the graph picture into mathematics, I think an easy way is the mean value theorem.
With it you find
$$n^s-(n-1)^s=s(n+\xi)^{s-1}\to 0$$

Excellent, nice and easy proof!