Lemma Let $( x,y)$ be a pair of positive rational numbers with $x >y$. $x^{y} =y^{x}$ hold if and only if $( x,y) =\left(( 1+1/n)^{n+1} ,( 1+1/n)^{n}\right)$ for $n\in \mathbb{N}$. proof of lemmaIt is not difficult to verify that $( x,y) =\left(( 1+1/n)^{n+1} ,( 1+1/n)^{n}\right)$ satisfies $x^{y} =y^{x}$. Suppose that $x^{y} =y^{x}$. Since $x/y >1$, we can take $r\in \mathbb{Q}^{+}$ such that $x/y=1+r$. Then we have $x=y^{x/y} =y^{1+r}$, which implise $1+r=y^{r} \ \cdots ( \twonotes )$. Let $r=m/n$ for $( m,n) \in \mathbb{N}^{2}$ with $\gcd( m,n) =1$. Since $y^{m/n} \in \mathbb{Q}$, we can take $y=z^{n}$ for $z\in \mathbb{Q}^{+}$. From $( 1)$ and $y=z^{n}$, we have $( m+n) /n=z^{m}$. So we can take $( A,B) \in \mathbb{N}^{2}$ such that $m+n=A^{m}$ and $n=B^{m}$. Thus we have $m=A^{m} -B^{m} \geq 2^{m} -1$. So we must have $m=1$. From $\twonotes $, we have $y=( 1+r)^{1/r} =( 1+1/n)^{n}$. From $x/y=1+r$, we have $x=( 1+1/n)^{n+1}$. $\blacksquare $ Let $( r_{1} ,\cdots ,r_{n}) \in \mathbb{Q}^{n}_{+}$ and $r_{n+j} =r_{j}$ for all $j\in \mathbb{N}$. We want to solve the equation : $r_1^{r_2}=r_2^{r_3}=...=r_{n-1}^{r_n}=r_n^{r_1}$ Suppose that $r^{r_{k+1}}_{k} =r^{r_{k+2}}_{k+1}$ for each $k\in \mathbb{N}$. WLOG, $\max\{r_{1} ,\dotsc ,r_{n}\} =r_{1}$. If $r_{i} =1$ for some $i$, then $r_{j} =1$ for all $j$. So we suppose, henceforth, that $r_{i} \neq 1$ for all $i$. If $r_{i} > 1$ for some $i$, then $r_{j} >1$ for all $j$. If $r_{i} < 1$ for some $i$, then $r_{j} < 1$ for all $j$. We consider the following two case $( 1) ,( 2)$ : $( 1)$ $r_{i} >1$ for all $i\in \mathbb{N}$. Since $r^{r_{2}}_{1} =r^{r_{k+1}}_{k}$ and $r_{1} \geq r_{k}$ for all $k\in \mathbb{N}$, $\min\{r_{1} ,\cdots ,r_{n}\} =r_{2}$. Since $r^{r_{3}}_{2} =r^{r_{k+1}}_{k}$ and $r_{k} \geq r_{2}$ for all $k\in \mathbb{N}$, $\max\{r_{1} ,\cdots ,r_{n}\} =r_{3}$. So we must have $r_{1} =r_{3}$. Suppose that $r_{i} =r_{i+2}$ for $i\in \mathbb{N}$. Since $r_{i} =r_{i+2}$ and $r^{r_{i+1}}_{i} =r^{r_{i+3}}_{i+2}$, we have $r_{i+1} =r_{i+3}$. So we must have $( r_{1} ,r_{2}) =( r_{2i-1} ,r_{2i})$ for all $i\in \mathbb{N}$. We consider the following two sub-cases $( 1a) ,( 1b)$ $( 1a)$ $n$ is an odd integer. Since $r_{n} =r_{1} =r_{n+1}$, we have $r_{1} =r_{i}$ for all $i\in \mathbb{N}$. Therefore, $( r_{1} ,\dotsc ,r_{n}) =( q,\dotsc ,q)$ for some $q\in \mathbb{Q}^{n}_{+}$. $( 1b)$ $n$ is an even integer, If $r_{1} =r_{2}$, then $( r_{1} ,\dotsc ,r_{n}) =( q,\dotsc ,q)$ for some $q\in \mathbb{Q}^{n}_{+}$. If $r_{1} \neq r_{2}$, by lemma, we have $\{r_{1} ,r_{2}\} =\left\{( 1+1/m)^{m+1} ,( 1+1/m)^{m}\right\}$ for $m\in \mathbb{N}$. Therefore, $r_{2i-1} =( 1+1/m)^{m+1}$ and $r_{2i} =( 1+1/m)^{m}$ or $r_{2i} =( 1+1/m)^{m+1}$ and $r_{2i-1} =( 1+1/m)^{m}$ for all $i\in \mathbb{N}$. $( 2)$ $r_{i} <1$ for all $i\in \mathbb{N}$. Since $r^{r_{2}}_{1} =r^{r_{k+1}}_{k}$ and $r_{1} \geq r_{k}$ for all $k\in \mathbb{N}$, $\max\{r_{1} ,\cdots ,r_{n}\} =r_{2}$. So we must have $r_{1} =r_{2}$. Suppose that $r_{i} =r_{i+1}$ for $i\in \mathbb{N}$. Since $r_{i} =r_{i+1}$ and $r^{r_{i+1}}_{i} =r^{r_{i+2}}_{i+1}$, we have $r_{i+1} =r_{i+2}$. So we must have $r_{1} =r_{i}$ for all $i\in \mathbb{N}$. Therefore, $( r_{1} ,\dotsc ,r_{n}) =( q,\dotsc ,q)$ for some $q\in \mathbb{Q}^{n}_{+}$. Hence we have the following conclusion : If $r\leq 1$ or $n$ is odd, then $( r_{1} ,\dotsc ,r_{n}) =( q,\dotsc ,q)$ for $q\in \mathbb{Q}^{n}_{+}$. If $r >1$ and $n$ is even, then $\{r_{1} ,r_{2}\} =\left\{( 1+1/m)^{m+1} ,( 1+1/m)^{m}\right\}$ for $m\in \mathbb{N}$ and $( r_{2i-1} ,r_{2i}) =( r_{1} ,r_{2})$ for all $i\in \mathbb{N}$ otherwise $( r_{1} ,\dotsc ,r_{n}) =( q,\dotsc ,q)$ for $q\in \mathbb{Q}^{n}_{+}$. On the other hand, we can easily verify that these n-tuples give solutions.