For parameters $a$ and $b$, both of which are $\omega(1)$, $T(n) = T(n^{1/a}) + 1$, and $T(b) = 1$. Then $T(n)$ is

To solve the recurrence $T(n) = T(n^{1/a}) + 1$ with base case $T(b) = 1$:

Apply the change of variables by taking the logarithm to base $b$: Let $n = b^{a^k}$, which implies:

Now rewrite the recurrence in terms of $k$:

Define $S(k) = T\left(b^{a^k}\right) = T(n)$. Then:

For the base case $n = b$, we have $a^k = 1 \implies k = 0$:

Unrolling the recurrence:

Thus, $T(n) = \Theta(\log_a \log_b n)$.

Therefore, option (A) is correct.