I would like to prove that the statement $40^n = O(2^n) $ is false
Would the following suffice as a proof?
Let k be some arbitrary number. Let c = $\frac {40^k}{2^k}$. Then if n>k
$\frac {40^n}{2^n}=\frac {40^k}{2^k}*\frac {40^{n-k}}{2^{n-k}}$.
Then it follows that $\frac {40^{n-k}}{2^{n-k}}>1$ and so we are going uphill.
If n>k,
$\frac {40^n}{2^n}>\frac {40^k}{2^k}=c$
and so
${40^n} \not \lt c(2^n)$