How to show that $\pi^\pi$ is not a integer? This goes without saying, but, I can't use a calculator to evaluate $\pi^\pi$. I think we need to find a integer $x$ such that
$$x<\pi^\pi < x+1. \tag{1}$$
However, since I have no ideia what $\pi^\pi$ looks like, probably I will not find $x$, but if I can prove that such integer $x$ exists, will be enough. But this seems like a difficult problem.
I can use a calculator for other things, for example: evaluating $\pi,\pi^2,e^{27}$  or $\log,  \sin$ etc. I tried taking the $\log$ base $\pi$ in $(1)$ to simplify $\pi^\pi$ to just $\pi$.
Maybe this approach is a wrong one.
This is not a "homework problem", is just something that I found it interesting to do and learn more.
I'll appreciate any insight and improved tags. Thanks!
Also, I don't think using a power series for $\pi^x$ is fair, because that's how calculators find the number in the first place. 

What about $\pi^{\pi^{{\pi}^{\pi}}}$ ? This is an open problem, so I wanted to see if there's a way to prove that $\pi^\pi$ is not integer in a more "analytical way" but also using mathematical softwares if needed, but thanks for the answers.

 A: The main question to answer is "How do you define $\pi$?" 
Once we know that, we can start to find upper and lower bounds for $\pi$ that show what your calculator showed you: That 
$$ 36 < \pi^{\pi} < 37. $$
A: First of all, notice that $(x^x)' = x^x(1+\ln(x))$ meaning that we need the precision of about at least $0.01$ for discussing the integer evaluation around $30$ where we expect the possible integer to land.
Take that
$$\sqrt{163} - \sqrt{67} \approx 4\ln(\pi)$$
(The constant is actually quite precise $4.0025$.)
Now again with quite some precision
$$e^{\pi \sqrt{163}} \approx 640320^3+744$$
$$e^{\pi \sqrt{67}}  \approx 5280^3+744$$
both very close to an integer due to Heegner numbers involved.
Now
$$\frac{e^{\pi \sqrt{163}}}{e^{\pi \sqrt{67}}}=e^{\pi (\sqrt{163}-\sqrt{67})} \approx e^{4 \pi \ln(\pi)}$$
Meaning
$$\pi^{4\pi} \approx \frac{640320^3+744}{5280^3+744} \approx \frac{640320^3}{5280^3} = \left ( \frac{1334}{11} \right)^3 \approx \left ( \frac{1331}{11} \right)^3 = 121^3=11^6$$
If $\pi^{\pi}$ is an integer so is $\pi^{2\pi}$ and we find it being close to $1331$.
However $\pi^{\pi} = \sqrt{11^3}=\sqrt{1331}$ cannot be an integer as there is no close square to $1331$, at least not within the error margin of the estimations involved.
(Notice that the method completely avoids dealing with big numbers.)
A: $$3.141 < π < 3.142$$
$${3.141}^{3.141} \approx 36.146 < 36.147 < π^{3.141} < π^π < π^{3.142} < {3.142}^{3.142} \approx 36.461$$
$$36.147 < π^π < 36.461$$
Therefore, $π^π$ is not an integer.
A: Archimedes proved directly from geometry (by inscribing and circumscribing regular 96-gons to a circle) that
$$
3+10/71<\pi<3+1/7.
$$
It is fairly straightforward to manipulate the inequalities to show that
$$
36<(3+10/71)^{3+10/71}\textrm{ and } (3+1/7)^{3+1/7}<37. 
$$
It reduces to comparing large integers, e.g. $31^{31}<7^{31}37^7$ for the second one. Since the function $x^x$ is monotone increasing for $x>1$ (by taking the derivative, or arguing from monotonicity of addition and multiplication for positive integers), $36 < π^π < 37$ is not an integer.
A: We have $\pi<\frac{22}7$. To see that $\pi^\pi<37$, we check that $(22/7)^{22/7}<37$, or equivalently
$$22^{22}<37^7\cdot 7^{22}. $$
This is something that can be done without calculator (though I won't):
$$ 341427877364219557396646723584<94931877133\cdot 3909821048582988049$$
It is possible to make a similar lower estimate of the same form, but that would require much higher powers - good luck. So in reality, a pocket calculator throws out $36.4621596$ and even if we allow an incredible 1% error in the calculation, this is safely between $36$ and $37$.
