To analyze the given condition, we need to look at the behavior of the function $f(x) = a^{2^x}$. The defining feature of an increasing function is that as $x$ increases, $f(x)$ also increases. For exponential functions like $f(x) = a^{2^x}$, the function's growth behavior is determined by the base of the exponent, which is $a^2$.

If the base $a^2$ is greater than 1, then the function is increasing. This happens in two separate cases:
1. When $a > 1$, then $a^2 > 1$.
2. When $a 1$ since squaring any real number always yields a positive result.

Therefore, the condition "$a > 1$" implies that the function $f(x) = a^{2^x}$ is increasing. However, it is not the only condition for the function to be increasing since $a 1$" is a sufficient condition for the function $f(x) = a^{2^x}$ to be increasing within its domain.

$$\boxed{\text{The correct answer is B: Sufficient condition}}$$