$\underset{n \to \infty}{\text{lim}} (1-p)^{n-d-1}$, given $p = o(n^{-(d+1)/d})$, and $d \geq 0$ a constant.

Question

This is a step in a proof I’m reading. I did a bunch of calculations and it seems like the answer is $1$. But I’m not sure so I’d like to get it checked, or to see a cleaner argument.

Also, my argument relies on the following: $$p = o(n^{-(d+1)/d}) \Rightarrow \frac{p}{n^{-(d+1)/d}} \to 0 \Rightarrow pn^{1 + (1/d)} = pnn^{(1/d)} \to 0.$$

So as $n \to \infty$, $p$ must tend to $0$ faster than both $n$ and $n^{(1/d)}$ (combined) tend to infinity. Is this a reasonable argument?