Correct my proof about : $x^{x^{x^{x^{x^{x}}}}}$ is convex for $0.25Let $0.25<x<0.27$ then define :
$$f(x)=x^{x^{x^{x^{x^{x}}}}}$$
Claim:
The second derivative of $f(x)$ is strictly positive .

Proof :
Let $0.25<x<0.27$ then define :
$$g(x)=x^{x^{x^{x}}}$$
$g(x)$ is convex .
Proof :
Fact 1:
On $(0.25,0.27)$ the function $k(x)=x^x$ is convex

Proof : Left to the reader using derivatives and logarithm .
Fact 2 :
On $(0.25,0.27)$ the function $l(x)=x^{x^x}$ is concave increasing

Proof :
Trick : we consider the function $l(x)=x^{k(x)}$ we have :
$$l''(x)=x^{k(x) - 2} (x^2 \ln(x) k''(x) + 2 x k'(x) + (x \ln(x) k'(x) + k(x))^2 - k(x))$$
Now we have as obvious things :
$$x^2 \ln(x) k''(x)<0$$
$$2xk'(x)<0$$
$$(x \ln(x) k'(x) + k(x))^2 - k(x)<\left(0.25\ln\left(0.25\right)k'\left(0.25\right)+k\left(0.25\right)\right)^{2}-k\left(0.27\right)<0$$
So the function is indeed concave and we deduce that $l(x)$ is increasing .
Fact 3:
Now we consider the function :
$$g(x)=x^{l(x)}$$
We have :
$$g''(x)=x^{l(x) - 2} (x^2 \ln(x) l''(x) + 2 x l'(x) + (x \ln(x) l'(x) + l(x))^2 - l(x))$$
A calculation gives :
$$l'(x)>1.15$$
So :
$$2xl'(x)>0.5\cdot 1.15$$
In the same vein we have :
$$2 x l'(x) + (x \ln(x) l'(x) + l(x))^2 - l(x)>2\cdot0.25\cdot1.15+\left(0.25\ln\left(0.25\right)l'\left(0.27\right)+l\left(0.27\right)\right)^{2}-l\left(0.27\right)>0$$
So the function $g(x)$ is convex .
We are done.
Fact 4 :
Now we introduce the function :
$$h(x)=x^{g(x)}$$
We have :
$$h''(x)=x^{g(x) - 2} (x^2 \ln(x) g''(x) + 2 x g'(x) + (x \ln(x) g'(x) + g(x))^2 - g(x))$$
As $g''(x)>0$ then $x^2 \ln(x) g''(x)<0$
We have again $g'(x)<0$
and It's not hard to show $(x \ln(x) g'(x) + g(x))^2 - g(x)<0$ using again particular values I mean maximize and minimize the different functions .
We deduce that on $0.25<x<0.27$ the function $x^{x^{x^{x^{x}}}}$ is concave and increasing.
Now introducing a similar function to $h(x)$ and using the same way the conclusion follow .
We are done !
Question :
Is it correct ?Have you any reference on the subject?Can we hope to have the general case even if we have the prohibition of calculus ?
 A: I have not found any errors in your proof, so I think that your proof is correct.
You have omitted some details in your proof, so in the following, I'm going to write a proof with the details you have omitted.
Claim : $x^{x^{x^{x^{x^{x}}}}}$ is convex for $0.25<x<0.27$.
Proof :

*

*Let $k(x):=x^x$. Then, $k''(x)=x^{x-1}\bigg(1 + x(1+\ln x)^2\bigg)\gt 0$, so $k'(x)=x^x(1+\ln x)$ is increasing with $k'(0.27)\lt 0$, which implies $k'(x)\lt 0$.


*Let $l(x):=x^{x^x}=x^{k(x)}$. Then, $l'(x)=\dfrac{l(x)}{x}\bigg(xk'(x)\ln x+k(x)\bigg)\gt 0$ and
$$l''(x)=x^{k(x) - 2} (x^2 \ln(x) k''(x) + 2 x k'(x) + (x \ln(x) k'(x) + k(x))^2 - k(x))$$ One gets $x^2 \ln(x) k''(x)<0$ and $2xk'(x)<0$. Also, let $F(x):=(x \ln(x) k'(x) + k(x))^2$. Then, $$F'(x)=2\underbrace{(x \ln(x) k'(x) + k(x))}_{\text{positive}}\underbrace{\bigg((\ln(x)+2)k'(x)+x\ln(x)k''(x)\bigg)}_{\text{negative}}\lt 0$$Since $F(x) - k(x)<F(0.25)-k\left(0.27\right)<0$, one obtains $l''(x)\lt 0$.


*Let $g(x):=x^{x^{x^x}}=x^{l(x)}$. Then, $$g''(x)=x^{l(x) - 2} (x^2 \ln(x) l''(x) + 2 x l'(x) + (x \ln(x) l'(x) + l(x))^2 - l(x))$$ Here, one can show $g''(x)\gt 0$ in an easier way than yours. Since $x^2 \ln(x) l''(x)+ (x \ln(x) l'(x) + l(x))^2\gt 0$ and $2xl'(x)-l(x)>2\cdot 0.25\cdot l'(0.27)-l(0.27)\gt 0$, one gets $g''(x)\gt 0$. Also, $g'(0.27)\lt 0$ implies $g'(x)\lt 0$.


*Let $h(x):=x^{x^{x^{x^{x}}}}=x^{g(x)}$. Then, $h'(x)=\dfrac{h(x)}{x}\bigg(xg'(x)\ln x+g(x)\bigg)\gt 0$ and $$h''(x)=x^{g(x) - 2} (x^2 \ln(x) g''(x) + 2 x g'(x) + (x \ln(x) g'(x) + g(x))^2 - g(x))$$One has $x^2 \ln(x) g''(x)+2xg'(x)<0$. Also, let $G(x)=(x \ln(x) g'(x) + g(x))^2$. Then, $$G'(x)=2\underbrace{(x \ln(x) g'(x) + g(x))}_{\text{positive}}\underbrace{\bigg((\ln(x)+2)g'(x)+x\ln(x)g''(x)\bigg)}_{\text{negative}}\lt 0$$Since $G(x) - g(x)\lt G(0.25) - g(0.27)<0$, one gets $h''(x)\lt 0$.


*Let $f(x):=x^{x^{x^{x^{x^x}}}}=x^{h(x)}$. Then, $$f''(x)=x^{h(x) - 2} (x^2 \ln(x) h''(x) + 2 x h'(x) + (x \ln(x) h'(x) + h(x))^2 - h(x))$$ One has $x^2 \ln(x) h''(x)+ (x \ln(x) h'(x) + h(x))^2\gt 0$. Also, one obtains $2xh'(x)-h(x)>2\cdot 0.25\cdot h'(0.27)-h(0.27)\gt 0$. Therefore, it follows from $f''(x)\gt 0$ that $f(x)$ is convex.$\quad\blacksquare$
A: I can't find anything wrong with your proof. It seems alright, as you can see in the graph as well.
Of course we can have a general case as is evident from your proofs. Just use induction to prove the general case.
The other question about references on this subject, there seems to be nothing online about convexity of the function. However, in case you don't know, the generalization of the function you are dealing with is known as tetration or power tower (i.e., repeated exponentiation). I did some amateur reading on this topic few months back, but I don't remember ever dealing with convexity of the function. Anyways, this function has got some nice properties that may help you with your problem, including an interesting result about convergence. It becomes more interesting in complex numbers. In case you want to really go deep into this, here is a detailed paper dealing with this function.
Hope that helps.
