For convenience we define $f(1) = 1$ and $f(n + 2) = (n + 2)^{f(n)}$ for odd positive integers $n$. Then the question asks for the last two digits of $f(19)$.

Since $f(13) = 13^{f(11)}$ is odd, we have $f(15) = 15^{f(13)} \equiv (-1)^{f(13)} = -1 \equiv 3 \pmod 4$.

As $f(17) = 17^{f(15)}$ and the units digits of the powers of 7 (also the powers of 17) follows the pattern 7, 9, 3, 1, 7, 9, 3, 1, which repeats itself every four terms, we conclude that the units digit of $f(17)$ is the same as that of $17^3$, which is 3.

Finally, if we look at the last two digits of the powers of 19, we will see the pattern 19, 61, 59, 21, 99, 81, 39, 41, 79, 01, 19, 61, which repeats itself every 10 terms. As $f(17)$ has units digit 3, the last two digits of $f(19) = 19^{f(17)}$ are the same as those of $19^3$, which are 59.