About $e^{\pi}\gt {\pi}^e, \ e^{e^{\pi}}\lt {\pi}^{{\pi}^{e}},e^{{\pi}^{{\pi}^e}}\gt {\pi}^{e^{e^{\pi}}}$ and their generalization Let us define a sequence $\{(a_n,b_n)\}$ as
$$(a_1,b_1)=(e,\pi),\ \  (a_{2n},b_{2n})=(e^{b_{2n-1}},{\pi}^{a_{2n-1}}),\ \ (a_{2n+1},b_{2n+1})=(e^{a_{2n}},{\pi}^{b_{2n}})$$
for $n=1,2,3,\cdots$.
Then, here is my question.
Question : Are the followings ture?
$$a_{4m}\gt b_{4m},\ \ a_{4m+1}\gt b_{4m+1},\ \ a_{4m+2}\lt b_{4m+2},\ \ a_{4m+3}\lt b_{4m+3}$$
for any $m\in\mathbb N$.
Examples : $(a_2,b_2)=(e^{\pi},{\pi}^{e}),\ (a_3,b_3)=(e^{e^{\pi}},{\pi}^{{\pi}^e}),\ (a_4,b_4)=(e^{{\pi}^{{\pi}^e}},{\pi}^{e^{e^{\pi}}}),\ (a_5,b_5)=({e}^{{e}^{{\pi}^{{\pi}^{e}}}},{\pi}^{{\pi}^{{e}^{{e}^{\pi}}}}).$
Motivation : We know $e^{\pi}\gt {\pi}^e, \ e^{e^{\pi}}\lt {\pi}^{{\pi}^{e}}$. I've been interested in the generalization of this inequalities. I've got the followings by using computer : 
$$e^{{\pi}^{{\pi}^e}}\gt {\pi}^{e^{e^{\pi}}},\ \ {e}^{{e}^{{\pi}^{{\pi}^{e}}}}\gt {\pi}^{{\pi}^{{e}^{{e}^{\pi}}}},\ \ {e}^{{\pi}^{{\pi}^{{e}^{{e}^{\pi}}}}}\lt {\pi}^{{e}^{{e}^{{\pi}^{{\pi}^{e}}}}},\ \ {e}^{{e}^{{\pi}^{{\pi}^{{e}^{{e}^{\pi}}}}}}\lt {\pi}^{{\pi}^{{e}^{{e}^{{\pi}^{{\pi}^{e}}}}}}.$$
Then, I reached the above expectation, but I'm facing difficulty. Can anyone help?
 A: I've just been able to prove the following theorem : 
Theorem : For any $m\in\mathbb N,$$$a_{4m}\gt b_{4m}, \ a_{4m+1}\gt b_{4m+1}, \ a_{4m+2}\lt b_{4m+2}, \ a_{4m+3}\lt b_{4m+3}.$$ 
In the following, suppose that we know that $e^{\pi}\gt {\pi}^e$ and that 
$$\begin{align}\pi=3.14\cdots,\ e=2.71\cdots\qquad(1)\end{align}$$
Lemma 1 : $1\lt \log\pi\lt\frac{\pi}{e}.$
Proof : Since $\pi\gt e$, we get $\log\pi\gt 1$. Also, since $e^{\pi}\gt {\pi}^{e},$ we get $\pi\gt e\log\pi$.
Lemma 2 : $(\log\pi)e^c\lt {\pi}^{c}$ for $c\gt 1$. 
Proof : Since $c\gt 1$, by lemma 1, $\log\pi\lt\frac{\pi}{e}\lt\left(\frac{\pi}{e}\right)^c.$ Then, all we need to do is to multiply $e^c$.
Lemma 3 : ${\pi}^{{\pi}^{{e}^{{e}^{x}}}}\lt{e}^{{e}^{{\pi}^{{\pi}^{y}}}}$ for $1\lt x\lt y.$
Proof : It's sufficient to prove $(\log\pi){\pi}^{{e}^{{e}^{x}}}\lt {e}^{{\pi}^{{\pi}^{y}}}$. Noting that $a=e^{\log a}$ for $a\gt 0$, by lemma2, 
$$(\log\pi){\pi}^{{e}^{{e}^{x}}}=(\log\pi)e^{(\log\pi)e^{{e}^{x}}}\lt (\log\pi)e^{{{\pi}^{{e}^{x}}}}\lt {\pi}^{{\pi}^{{e}^{x}}}\lt {\pi}^{{\pi}^{{e}^{y}}}={\pi}^{e^{(\log\pi)e^{y}}}\lt {\pi}^{e^{{\pi}^{y}}}=e^{(\log\pi)e^{{\pi}^{y}}}\lt e^{{\pi}^{{\pi}^{y}}}$$
Now we are going to prove the theorem.
Proof for theorem : First, let us prove $a_{4m+1}\gt b_{4m+1}$ for $m\in\mathbb N$ by induction. In order to prove the $m=1$ case, it is sufficient to prove 
$$\begin{align}{\pi}^{{\pi}^e}\gt e^{e^{\pi}}(\log\pi)+\log(\log\pi)\qquad(2)\end{align}$$
Now let us consider the following equation : 
$$\begin{align}\log\left(\frac{{\pi}^{{\pi}^{e}}}{e^{e^{\pi}}(\log\pi)}\right)={\pi}^e(\log\pi)-e^{\pi}-\log(\log\pi)\qquad(3)\end{align}$$
Now let us define a function $f(x)$ as the following : 
$$f(x):=x^ee^{-x}(\log x)\ \ (x\in [e,\pi])$$
Hence we get
$$f^{\prime}(x)=x^{e-1}e^{-x}\{1-(x-e)\log x\}.$$
By mean value theorem, there exists $\xi\in(e,\pi)$ such that
$$\begin{align}f(x)-f(e)=(\pi-e)f^{\prime}(\xi)\qquad(4)\end{align}$$
Here, noting that $f(\pi)={\pi}^ee^{-\pi}(\log\pi)$ and that $f(e)=1$, $(4)$ tells us that 
$${\pi}^e\log\pi-e^{\pi}=e^{\pi}(\pi-e)e^{-\xi}{\xi}^{e-1}\{1-(\xi-e)\log\xi\}\gt(\pi-e)\cdot 1\cdot e^{e-1}\{1-(\pi-e)\log\pi\}.$$
Here, since $0\lt \pi-e\lt 3.2-2.7=\frac 12$, lemma 1 tells us that 
$${\pi}^e\log\pi-e^{\pi}\gt e\left(\frac{\pi}{e}-1\right)\cdot 2\left(1-\frac 12\cdot\frac{\pi}{e}\right)=\left(\frac{\pi}{e}-1\right)(2e-\pi)\gt (\log\pi-1)(2e-\pi).$$
Then, noting that $\log\pi-1=e^{\log(\log\pi)}-1\gt (1+\log(\log\pi))-1=\log(\log\pi)$ and that $2e-\pi\gt 2\times 2.7-3.2=2.2\gt 2$, we get
$$\begin{align}{\pi}^e\log\pi-e^{\pi}\gt 2\log(\log\pi)\qquad(5)\end{align}$$
Hence, $(3)(5)$ tells us that
$$\log\left(\frac{{\pi}^{{\pi}^e}}{e^{e^{\pi}}(\log\pi)}\right)\gt 2\log(\log\pi)-\log(\log\pi)=\log(\log\pi).$$
Hence, since 
$${\pi}^{{\pi}^{e}}\gt e^{e^{\pi}}(\log\pi)e^{\log(\log\pi)}\gt e^{e^{\pi}}(\log\pi)(1+\log(\log\pi))\gt e^{e^{\pi}}\log\pi+\log(\log\pi),$$
we know that the $m=1$ case is proven.
Now suppose that $a_{4m+1}\gt b_{4m+1}$ for some $m\ge 1$. Then, lemma 3 tells us that 
$${\pi}^{{\pi}^{e^{e^{b_{4m+1}}}}}\lt e^{e^{{\pi}^{{\pi}^{a_{4m+1}}}}}.$$
Hence, since 
$$b_{4m+5}={\pi}^{b_{4m+4}}={\pi}^{{\pi}^{a_{4m+3}}}={\pi}^{{\pi}^{e^{a_{4m+2}}}}={\pi}^{{\pi}^{e^{e^{b_{4m+1}}}}},$$
we get $b_{4m+5}\lt a_{4m+5}$. 
Now we can prove that $a_{4m+1}\gt b_{4m+1}$ for any $m\in\mathbb N$.
Noting that $\log\pi\gt 1$, we get
$$a_{4m}=\log e^{a_{4m}}=\log a_{4m+1}\gt \log b_{4m+1}=\log {\pi}^{b_{4m}}=b_{4m}\log\pi\gt b_{4m}.$$
Also, we get
$$a_{4m+2}=e^{b_{4m+1}}\lt {\pi}^{b_{4m+1}}\lt {\pi}^{a_{4m+1}}=b_{4m+2},$$
and 
$$a_{4m+1}=e^{a_{4m+2}}\lt {\pi}^{a_{4m+2}}\lt {\pi}^{b_{4m+2}}=b_{4m+3}.$$
Hence, we now know that the proof of theorem is completed.
