Question involving exponential tower of 19 Consider:
$$
y = \underbrace{19^{19^{\cdot^{\cdot^{\cdot^{19}}}}}}_{101 \text{ times}}
$$
with the tower containing a hundred $ 19$s. Take the sum of the digits of the resulting number. Again, add the digits of this new number and get the sum. Keep doing this process till you reach a single-digit number. Find the number.
Here's what I tried so far:- Every number which is a power of $19 $ has to end in either $1 $ or $ 9$. Also, by playing around a bit, taking different powers of $19$, I found that regardless of the power of $19$, whether it is an odd or an even number, the single-digit number obtained at the end is always $ 1$. I've been trying to prove this, but I've no idea on how to do it. Can anyone help me out?

$$10^0a_0+10^1a_1+10^2a_2+\ldots+10^na_n=(a_0+a_1+\ldots+a_n)+\text{a multiple of }9.$$
Therefore taking the sum of the digits of a number gives you a number that leaves the same remainder, in the division by $9$, as the one you had before (and it is also smaller as long as the original number is not $<10$). Therefore the process ends with a one-digit number that leaves the same remainder under division by $9$ are your number.
