Twice and thrice tetration

Relevant wiki: Finding the last few digits of a power

The last two digits of the powers of 9 repeats with the cycle length of 10.

Relevant wiki: Carmichael's Lambda Function

Let $N = 9^{9^{9^9}}$ . Since $\gcd(9,100) = 1$ , we can apply Euler's theorem and the related Carmichael lambda as follows.

$\large \begin{aligned} N & \equiv 9^{9^{9^9}\bmod \color{#3D99F6} \lambda(100)} \text{ (mod 100)} & \small \color{#3D99F6} \text{Carmichael lambda }\lambda (100)=20 \\ & \equiv 9^{9^{9^9 \bmod \color{#D61F06} \lambda (20)}\bmod \color{#3D99F6} 20} \text{ (mod 100)} & \small \color{#D61F06} \text{Again }\gcd(9,20) = 1 \\ & \equiv 9^{9^{9^{9 \color{#3D99F6} \bmod 2} \bmod \color{#D61F06}4}\bmod 20} \text{ (mod 100)} & \small \color{#3D99F6} \text{and }\gcd(9,4) = 1 \\ & \equiv 9^{9^{9^{\color{#D61F06}1} \bmod 4}\bmod 20} \text{ (mod 100)} \end{aligned}$

Therefore, $9^{9^{9^9}} \equiv 9^{9^9} \text{ (mod 100)}$ . Yes , the two numbers have the same last two digits.

Note that for any $n\geq 0$ , by the binomial theorem, $9^n = (10-1)^n = \sum_{k=0}^n \binom{n}{k} 10^k (-1)^{n-k} \equiv (-1)^n (1-10n) \pmod{100}$ In particular, $9^{10} \equiv (-1)^{10} (1-10^2) \equiv 1 \pmod{100}$

This means that $9^a, 9^b$ have the same last two digits if $a\equiv b \pmod{10}$ , and in this case, both the exponents have the form (for $m=1$ or $m=9$ ) $9^{9^m} \equiv (-1)^{9^m} \equiv -1 \pmod{10}$ so we can conclude that both numbers have the same last two digits.

Note that $9^2\equiv81\equiv1\pmod{40}$ , so no matter $9^9$ or $9^{9^9}$ wil leave remainder $9\pmod{40}$ , as $\phi(100)=40$ , we know both wil congruent $9^9\pmod{100}$ which gives the $\boxed{SAME}$ last two digits.

$9^{9}=387420489$ , which is notable for ending in $89$ .

The last two digits of $9^{n}$ have a cycle of length 10, meaning $9^{a}$ ends in the same two digits as $9^{\text{last digit of a}}$ .

So $9^{9^{9}}$ ends with $89$ and since this ends in $9$ , then $9^{9^{9^{9}}}$ also ends in $89$

This is not level 1 ; this needs congruence knowledge.

Rewrite the question as: $((10-1)^{{9}^9}$ and $((10-1)^{{{9}^9}^9}$

By the binomial expansion: $(10-1)^{x} =\displaystyle \sum_{k=0}^x {x \choose k }10^{x-k} (-1)^{k}$

For the last components of the expansion, we get: $=...+ (10)^{x-x} (-1)^{x}=...+(-1)^{x}$

Hence, the last component will either be: $(-1)^{odd}=-1$ or $(-1)^{even}=1$

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Now consider all the components before this last one, i.e.:

$\displaystyle \sum_{k=0}^{x-1} {x \choose k }10^{x-k} (-1)^{k}={x \choose 0 }10^{x}+{x \choose 1 }10^{x-1}(-1)^{-1}+{x \choose 2 }10^{x-2}(-1)^{-2}+{x \choose 3 }10^{x-3}(-1)^{-3}+... ={x \choose 0 }10^{x}-{x \choose 1 }10^{x-1}+{x \choose 2 }10^{x-2}-{x \choose 3 }10^{x-3}+...$

This will be a number $n$ such that $n=k \times 10$ , i.e. a multiple of 10 or a number whose last digit is 0.

Hence, when we add the last component (either $-1$ or $1$ as shown before), it will either:

Subtract $-1$ , i.e. $(k \times 10 -1)$ , making the last digit $9$

or

Add $1$ , i.e. $(k \times 10 +1)$ , making the last digit $1$

Regardless, the last digit is always odd which makes the number always odd .

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Now consider: $((10-1)^{{9}^9} = ((10-1)^{{(10-1)}^{9}} =((10-1)^{{(10-1)}^{odd}} =((10-1)^{odd}$ which has last digit $9$ as shown before

and:

$((10-1)^{{9}^{{9}^{9}}} = ((10-1)^{{(10-1)}^{{(10-1)}^{9}}} = ((10-1)^{{(10-1)}^{{(10-1)}^{odd}}} = ((10-1)^{{(10-1)}^{odd}} = ((10-1)^{odd}$ which has last digit $9$ as shown before.

Therefore, the last digit of both numbers is 9.

The power of 9 will end in either 1 (when the exponent is even) or 9 (when the exponent is odd)... ... ... (1) At the same time any power of 9, ending in either 1 or 9, will be odd... ... ... (2) Here 9 is raised to another power of 9. So in both cases the exponent is odd, according to (2). So, according to (1), both will end in 9.

Kinda can't find that simple solution in other answers: $9 \times (10n + 9) = 90n + 80 + 1$ give $m = 9n + 8$ $9 \times (10m + 1) = 90m + 9$ give $n = 9m$
repeat to the first equation
This goes on like a loop. If you start with 9=9 and 9 square = 81 you can easily prove that odd powers of 9 display 9 as digits unit while even powers have 1. Since both powers of 9 were clearly odd powers (any power of an odd number is still an odd number and vice versa) the units digit of both should be 9. Sorry for messy English, not my first language...

My thought process: The question is very specific so the answer can't possibly be "no". Voila :D

I am from china. I am honored to introduce my methods to you. I think it is very simple and easy to understand. Let's go straight to the end.9^1---09 9^2----81 9^3-----29 9^4------61 and so on . 9^n-------???? If the"n" is an odd number the second inverse is satisfied --- {0 2 4 6 8 0 2...},and the first inverse must be nine and the reciprocal second number is "n-1"---------- If the "n"is an even number --------{8 6 4 2 0 8 6...},and the first inverse must be "1" So,9^9 is an odd number 9^9^9------------89 9^(9^9^9) is an odd number and the"n"------is "9" so------89