(Note: I'm reposting this, as I posted the original too late in the evening to gain anyone's notice.)
A contest problem (#2 on the 2010 Virginia Tech Math Competition) proffers the solver the functions $f_i(m)$, where $f_1(m) = m$ and $f_{i+1}(m) = {f_i(m)}^{f_i(m)}$, and asks them to deduce the remainder of $f_{100}(75)$ modulo $n = 17$.
I believe I've solved this one. Computations show that $(75^{75})^{(75^{75})} \equiv 75 \mod(17)$, and from this the laws of exponents yield the same congruence modulo $16$, and also modulo $\varphi(16) = 8$. (This is because $(75^{75})^{(75^{75})} = 75^{(75)(75^{75})}$, so $(75)(75^{75}) \equiv 1 \mod(16)$ by Euler-Fermat, and thus, exponentiating $75$ to each side, $(75^{75})^{(75^{75})} \equiv 75 \mod(16)$ - because $(75)(75^{75}) = 1 + k\varphi(16)$ - and therefore modulo $8$ also.) These congruences, taken together, can be pushed forward to yield
$$f_{i+2}(75) \equiv f_{i}(75) \mod(17)$$
which of course solves the problem. (The answer is $14$.)
Another one that recently came up was showing $7^{7^{7^7}} \equiv 7^{7^7} \mod(10).$ (Here exponentiation is right-associative.) Since $7^4 = 2401 \equiv 1 \mod(10)$ and $\mod(4)$, and since $7^7 \equiv 7 \mod(4)$, $7^{(7^7)} = 7^{7 + 4k} \equiv 7^7 \mod(4)$, so
$$ 7^{(7^{(7^7)})} = 7^{(7^7 + 4k)} \equiv 7^{7^7} \mod(10). $$
My question: what are some general rules for handling these kinds of problems? The ad-hoc approach is too harrowing, but trying some different values of $m$ and $n$ yields not-obviously patterned behavior. Are there any theorems or techniques available? (For example, the generalization of Euler-Fermat in Qiaochu's answer here is nice.)
