Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$

If $a = 0$ then $s = 0$ satisfies $s \ge 0$ and $s^n = 0 = a$, since $0^n = 0^{n-1} \cdot 0 = 0$ for $n \ge 1$; so existence holds in that case and we assume $a > 0$ from here on.

Uniqueness holds as soon as a root exists: if $u, v \ge 0$ satisfy $u^n = v^n$, then strict monotonicity of $x \mapsto x^n$ on the nonnegatives rules out both $u < v$ and $v < u$, so $u = v$ by trichotomy.

Define $S = \{t \in \mathbb{R} : t \ge 0 \text{ and } t^n \le a\}$; then $0 \in S$, because $0 \ge 0$ and $0^n = 0 \le a$, so $S$ is nonempty.

The element $M := 1 + a$ is an upper bound of $S$: since $a > 0$ and $1 > 0$ we have $M > 1$ and $M > a$, so any $t > M$ satisfies $t > 1$, whence $t^n \ge t > M > a$ and $t \notin S$.

By the least-upper-bound property $s := \sup S$ exists in $\mathbb{R}$; moreover $s \ge 0$ because $0 \in S$, and $s \le M$ because $M$ is an upper bound and $s$ is the least one.

Put $B := M + 1 = a + 2$; then $B > M \ge 1 > 0$, so $B^{\,n-1} > 0$ and $n B^{\,n-1} > 0$, and every $x$ with $0 \le x \le M + 1$ satisfies $0 \le x \le B$.

Assume, for contradiction, that $s^n \ne a$; by trichotomy either $s^n < a$ or $s^n > a$.

(Case $s^n < a$.) Put $\delta := \min\{1, (a - s^n)(n B^{\,n-1})^{-1}\}$, which is $> 0$ since $a - s^n > 0$ and $n B^{\,n-1} > 0$, and put $h := 2^{-1}\delta$, so that $0 < h < \delta \le 1$ and $h < (a - s^n)(nB^{\,n-1})^{-1}$; then $0 \le s \le s + h \le M + 1 = B$, so the Lipschitz estimate gives $(s+h)^n - s^n \le nB^{\,n-1} h < a - s^n$, hence $(s+h)^n < a$ and $s + h \in S$, while $s + h > s$ contradicts that $s$ is an upper bound of $S$.

(Case $s^n > a$.) Here $s > 0$, since $s = 0$ would give $s^n = 0 \le a$; put $\delta := \min\{s, (s^n - a)(nB^{\,n-1})^{-1}\} > 0$ and $h := 2^{-1}\delta$, so that $0 < h < s$ and $h < (s^n - a)(nB^{\,n-1})^{-1}$; then $0 \le s - h \le s \le B$, so the Lipschitz estimate gives $s^n - (s-h)^n \le nB^{\,n-1}h < s^n - a$, hence $(s-h)^n > a$; applying the epsilon characterisation with $\varepsilon = h$ produces $t \in S$ with $t > s - h \ge 0$, whence $t^n > (s-h)^n > a$ by strict monotonicity, contradicting $t \in S$.

Both cases of the disjunction in step 3.2 are impossible, so the assumption fails and $s^n = a$; this $s$ is the unique nonnegative $n$-th root of $a$ by step 1.2, it satisfies $s > 0$ when $a > 0$ because $s = 0$ would force $a = s^n = 0$, and at $n = 1$ the element $a$ itself is a nonnegative solution of $s^1 = a$, so $a^{1/1} = a$; writing $a^{1/n} = \sqrt[n]{a}$ for it, the case $n = 2$ recovers the already published $\sqrt{a}$ of Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$.