Binomial expansion for a tower of powers or maybe not

$\begin{aligned} 35789^{99^{100^{1000^{1000}}}} & \equiv 89^{99^{100^{1000^{1000}}}} \text{ (mod 100)} & \small \blue{\text{Since }\gcd(89, 100)=1 \text{, Euler's theorem applies.}} \\ & \equiv 89^{99^{100^{1000^{1000}}}\bmod \lambda (100)} \text{ (mod 100)} & \small \blue{\text{Carmichael's lambda function }\lambda(100) = 20} \\ & \equiv 89^{99^{100^{1000^{1000}}}\bmod 20} \text{ (mod 100)} \\ & \equiv 89^{(100-1)^{100^{1000^{1000}}}\bmod 20} \text{ (mod 100)} \\ & \equiv 89^1 \text{ (mod 100)} \\ & \equiv \boxed {89} \text{ (mod 100)} \end{aligned}$

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References:

• Euler's theorem
• Carmichael's lambda function

${{{35789^{99}}^{100}}^{1000}}^{1000} \equiv {35789^{(100-1)}}^{10^k} \equiv 35789^{a(100^m)+1} \equiv [((35790-1)^{100^m})^a](35789)\equiv 35789 \equiv 89 \ (\mod 100)$ where the term $((35790-1)^{100^m})^a$ will be $1(\mod 100)$ after the expansion.

Mod $100, 35789^x$ is congruent to $(-11)^x$ . If the unit's place digit of the power of $11$ be $n$ , then the number formed by the last two digits of the number is $10n+1 : 11^{10a+n}=100b+10n+1$ where $0\leq a, b\leq 9$ are positive integers. The last digit of $99^{2m}$ is $1$ , where $m$ is a positive integer. So the number formed by the last two digits of the given expression is $11$ , and hence the required answer is $100-11=\boxed {89}$ .