Proof that $x>0$ and $r,s\in\mathbb R\implies(x^{r})^{s}=x^{rs}$ I want to  prove the identity $(x^{r})^{s}=x^{rs}$ for real exponents and positive base.
My problem essentially boils down to this: I don't know how to prove that for $r_n,s_n\in\mathbb{Q},$ $$\lim_{n\to\infty}{\lim_{k\to\infty}{x^{r_ks_n}}}=\lim_{n\to\infty}{x^{r_ns_n}}.$$
Some of the results that are available for the proof:

*

*The other exponent identites $x^rx^s=x^{r+s}$, and $x^ry^r=(xy)^r$ for real numbers

*The corresponding identity (and all other exponent equalities and inequalities) for rational exponents

*That every real number can be expressed as the limit of some increasing (or decreasing if needed) sequence of rational numbers

*$x^r$ defined as the limit of $x^{r_n},$ where $r_n$ is a sequence of rationals that converges to $r$

*All the limit laws, including taking rational exponents inside and out of limits (not real ones of course)

Any help would be much appreciated. Please let me know if you need further clarifications.
 A: I suppose that you have already proven that the definition is consistent, i.e. for every pair of rational sequences $(r_n)_{n\geq1},(r'_n)_{n\geq1}$ that converge to the same value $r$ it follows that $\lim_n x^{r_n}=\lim x^{r'_n}$ for every $x>0$. With this in mind, it suffices to show that there exist rational sequences $(r_n)_{n\geq1}\to r$ and $(s_n)_{\geq1}\to s$ such that $\lim_n (x^{r})^{s_n}=\lim_n x^{r_n s_n}$.
$\color{#FF8700}{\bf{Lemma\ 1.}}$ If $r,s\in\mathbb{Q}$ and $x>0$ then $(x^r)^s=x^{rs}$.
Proof. It is a direct consequence of the well-known facts $(x^n)^m=x^{nm}$ and $\sqrt[m]{\sqrt[n]{x}}=\sqrt[nm]{x}$ for $n,m\in\mathbb{N}^*$.
$\color{#FF8700}{\bf{Lemma\ 2.}}$ For every rational sequence $(s_n)_{n\geq1}\to s\in\mathbb{R}$, there exists a rational sequence $(r_n)_{\geq1}\to r\in\mathbb{R}$ such that $\lim_n[(x^{r_n})^{s_n}-(x^r)^{s_n}]=0$ for a given $x>0$.
Proof. Take any rational sequence $(r'_k)_{k\geq1}\to r$ and notice that, since $\lim_k x^{r'_k}=x^r$ by definition, we have that $\forall n\geq1:\lim_k (x^{r'_k})^{s_n}=(x^r)^{s_n}\Rightarrow \exists (r'^{(n)}_{k_m})_{m\geq1}\subseteq (r'_k)_{k\geq1}:|(x^{r'^{(n)}_{k_m}})^{s_n}-(x^r)^{s_n}|<\frac{1}{n}$. Now consider $(r_n)_{n\geq1}=(r'^{(n)}_{k_n})_{n\geq1}$ noticing that it satisfies $\forall n\geq1:|(x^{r_n})^{s_n}-(x^r)^{s_n}|<\frac{1}{n}$ (so $\lim_n[(x^{r_n})^{s_n}-(x^r)^{s_n}]=0$) and that, because $r'^{(n)}_{k_n}$ is a term of the original sequence that appears somewhere after $r'_n$, $(r_n)_{n\geq1}\to r$ as well.
$\color{#FF6000}{\bf{Proposition.}}$ If $r,s\in\mathbb{R}$ and $x>0$ then $(x^r)^s=x^{rs}$.
Proof. By the previous lemma, for a given rational sequence $(s_n)_{n\geq1}\to s$, we can find another rational sequence $(r_n)_{n\geq1}\to r$ such that $((x^{r_n})^{s_n}-(x^{r})^{s_n})_{n\geq1}\to 0$. Now, since $((x^{r})^{s_n})_{n\geq1}\to (x^r)^s$ and $\lim_n[(x^{r_n})^{s_n}-(x^r)^{s_n}]=0$, we know that $(x^{r_n s_n})_{n\geq1}=((x^{r_n})^{s_n})_{n\geq1}\to (x^{r})^s$ where the first equality is a direct consequence of Lemma 1. But $(x^{r_n s_n})_{n\geq1}\to x^{rs}$. So $x^{rs}=(x^r)^s$.
