Prove $x \geq 2$ implies $x^{n} \geq 2^{n}$

Question

Prove $x \geq 2$ implies $x^{n} \geq 2^{n}$.

By induction. Clearly it holds for $n = 1$ by the assumption. Now assume $x \geq 2$ and $x^{k} \geq 2^{k}$ for some $k \in \mathbb{N}$.

Then combine the assumption inequality and $x \geq 2$ to get $x \cdot x^{k} \geq 2 \cdot 2^{k}$, which gives our result.

Is it correct?

Answer 1

Functions $F(x) = A^x$ are always increasing as long as $A > 1$. Thus, with $A=x$ and $x=n$, we have

$$ x \geq 2 \implies x^n \geq 2^n $$

Answer 2

\begin{align} x^{k+1} &=x(x^k) \\ &\ge x(2^k) \text{, since } x>0.\\ &\ge 2(2^k)\text{, since } x\ge2 \text{ and } 2^k >0\\ &\ge 2^{k+1} \end{align}