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The proof can be using calculus or finding a $n_o$ from which it holds that $c^n>n^k$ for $k \ge 1$ and $c>1$

Cate
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    If you want to prove that $c^n>n^k$ for all $k\geq1$, $c>1$ and $n\in\mathbb{N}$ then there's a problem since that's not true, however if you want to prove that for all $k\geq1$, $c>1$ we have that eventually $c^n>n^k$ then you can do so by computing $\lim_{x\to\infty} \frac{c^x}{x^k}$ and for that you can use L'Hopital's rule $k$ times. – Ignacio Henríquez May 28 '22 at 16:19
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    I was going to suggest the same as @IgnacioHenríquez. Another approach is: $c^n>n^k \iff n\ln(c) > k\ln(n) \iff n/\ln(n) > k/\ln(c)$ where the last term is constant. – David P May 28 '22 at 16:22
  • Thank you both! – Cate May 28 '22 at 17:01

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