True/False: $a\cdot n^k-b\cdot n^{k-1}=\Omega(n^k)$

Question

I need to prove or provide a counter example to this statement:

$a\cdot n^k-b\cdot n^{k-1}=\Omega(n^k)$ while $a,b>0$ and $k \ge 1$.

I think that this one is true, so I want to find $c>0$ and $n_0 \in \mathbb{N}$ such that: $0 \le c \cdot n^k \le a\cdot n^k-b\cdot n^{k-1}$.

I can say that $a\cdot n^k-b\cdot n^{k-1} \ge a\cdot n^k-b\cdot n^k=(a-b) \cdot n^k$.

I can't take $c=(a-b)$ because I don't know if $a>b$...

Any help?

Thanks!