My lecturer has told us that if $a > b, n^a \neq O(n^b)$ and i am struggling to prove this.
I have the following and would appreciate a hand to get to the end: Assume $n^a=O(n^b)$ for $a>b$, hence there must exist constants $c,n_0> 0$,such that $\forall n > n_0, n^a \leq cn^b$.
This tells us that $\frac{n^a}{n^b} = n^{a-b} \leq c$. As $a>b, a-b > 0$ and $n^{a-b} > 1$
From here, i know we must find a contradiction, maybe by choosing a value of c in terms of n, but i get lost here. Any help would be appreciated.