prove change in total probability of success in binomial distribution

A binomial distribution of $n$ samples and probability of success $p$ is defined as

$ P(k) = binom{n}{k} cdot p^kq^{n-k} $.

For a given value of $r$ where $r in mathbb{N} quad text{and } r gt 1$, if we decrease the probability of success by $hat{p} = frac{p}{r}$ and increase the samples $hat{n} = nr$, I want to prove

$P(kge 1) – P(hat{k}ge 1) gt 0 $

$Rightarrow left(1-P(k= 0)right) – left(1-P(hat{k}= 0)right) gt 0$

$Rightarrow P(hat{k}= 0) – P(k= 0) gt 0$

$Rightarrow binom{nr}{0} cdot hat{p}^0hat{q}^{nr} – binom{n}{0} cdot p^0q^{n} gt 0$

$Rightarrow left(1-hat{p}right)^{nr} – left(1-pright)^{n} gt 0$

$Rightarrow left(1-frac{p}{r}right)^{nr} – left(1-pright)^{n} gt 0$

I’m able to show it for specific examples but unfortunately unable to prove the last inequality in generic form.