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Let's start by the first simple thing, for any positive integer $a$ and for any positive integers $q$ we have: $$\forall k \ \ \ \ \ a\equiv a-kq \mod q $$ taking this to the next level we have: $$\forall k \ \ \ \ \ a^{a}\equiv a^{a-k\varphi(q)} \mod q $$ and so on :$\displaystyle$ $$\forall k \ \ \ \ \ a^{\displaystyle a^{a}}\equiv a^{\displaystyle ...


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The wrong thing is that $3\uparrow\uparrow\uparrow\uparrow3\neq3\uparrow\uparrow(3\uparrow\uparrow3)$ Hyperoperation (from tetration and so on) written in Knuth's notation satisfy the relation: $a\uparrow^nb=a\uparrow^{n-1}a\uparrow^{n-1}a\uparrow^{n-1}\dots \uparrow^{n-1}a$ where $n$ is the number of arrows, and $\uparrow^{n-1}$ is iterated b times. So ...



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