I'm trying to prove, as a lemma for a larger result, that when $n \geq 2$, $n > \sqrt{n}$.
I first tried induction. I don't think there's a rigorous way to check the base case other than to note that $2 > \sqrt{2} \approx 1.4142 \ldots$. If this result holds for some $k \geq 2$, then because $k > \sqrt{k}$, $k + 1 > \sqrt{k} + 1$. I need to show that $k + 1 > \sqrt{k+1}$, but I can't figure out how to change the right-hand side from a sum of squar roots (with $1 = \sqrt{1}$) instead a square root of sums without assuming the conclusion.
I also tried defining $f: \mathbb{N} \to \mathbb{N}$ by $f(n) = n - \sqrt{n}$ and showed that for a general $n$, $f(n) > 0$ if and only if $n \geq 2$. The steps are not however reversible, because I cannot assert that $n - \sqrt{n} > 0$ if and only if $(n - \sqrt{n})^2 > 0$. The forward direction certain holds, but there's no guarantee that the backward direction holds: that would require assuming the conclusion.
With that, I'm stuck. Some guidance and hints would be appreciated.