Let's determine the last two digits of the number

$$
7^{9^{9^{9}}}
$$
A product of a number of the form $4k+1$ and another number of the same form: $(4k+1)(4l+1) = 16kl + 4k + 4l + 1$, is a number of the same form. This implies that every power of $4k+1$ is of the same form, so $9^{9^9} = (4 \cdot 2 + 1)^{9^9} = 4k + 1$.

$$
7^{9^{9^9}} = 7^{7^{k+1}} = 7 \cdot (7^4)^k = 7 \cdot 2401^k
$$

But a number ending in 01, when multiplied by another number ending in 01:

$(100m+1)(100n+1) = 10000mn + 100m + 100n + 1 = (100mn + m + n) \cdot 100 + 1$, also results in a number ending in 01, so $2401^k$ ends in 01, and thus the last two digits of the desired number are: 07.

Lábos Elemér (Sátoraljaújhely, Kossuth g. IV. o. t.)

Note: The vast majority of solvers referred to the binomial theorem.