Step 1: Find the derivative of the function. Given $f(x) = x^{100} + \sin x - 1$, differentiating with respect to $x$, we get $f'(x) = 100x^{99} + \cos x$.

Step 2: Analyze the sign of $f'(x)$ in interval (A) $(0, 1)$ and (C) $(0, \pi/2)$. For $x > 0$, the term $100x^{99}$ is positive. In the interval $(0, \pi/2)$, $\cos x$ is positive. Therefore, $f'(x) > 0$ in these intervals, implying the function is increasing.

Step 3: Analyze the sign of $f'(x)$ in interval (B) $(\pi/2, \pi)$. Here, $x > \pi/2 > 1$, so $x^{99} > 1$, making $100x^{99}$ a large positive number (greater than 100). While $\cos x$ is negative in this interval, its magnitude is at most 1. Thus, $100x^{99} + \cos x$ is positive.

Step 4: Conclude the nature of the function. Since $f'(x) > 0$ in all the intervals (A), (B), and (C), the function is increasing in all these intervals. Therefore, the function is decreasing in none of these intervals.