Let \(b > 1\) be a real number and let \(C > 0\) be a constant. We want to show that there exists a natural number \(n\), for which \(b^n > C\).
Set \(x:= b~- 1\). Since \(x > 0\), we can use Bernoulli's inequality and conclude that
\[b^n=(1+x)^n \ge 1 + nx\]
for all \(n\ge 2\). Together with the Archimedean axiom, it follows that there exists a natural number \(n\in\mathbb N\) with \[nx > C - 1.\] Combining both inequalities, we get for this \(n\) that \(b^n > C\).