I tried to take to bound it by taking the limit of the geometric series but it didn't help. I also tried to write: $\sum\limits_{k=1}^{p}\frac{1}{(n+1)^{k}}< \sum\limits_{k=1}^{p}\frac{1}{n+1}=\frac{p}{n+1}$ which doesn't tell me anything I don't already know.
Thank you.
EDIT: I think I thought too fast, does this work?
$\sum\limits_{k=1}^{p}\frac{1}{(n+1)^{k}}< \sum\limits_{k=0}^{\infty }\frac{1}{(n+1)^{k}}-1=\frac{1}{1-\frac{1}{n+1}}-1=\frac{n+1}{n}-1=\frac{1}{n}$
Notice that this is merely a geometric sum, so we have
$$\sum_{k=1}^pr^k=r\frac{1-r^p}{1-r}$$
Take $r=\frac1{n+1}$ and notice that $1-r^p<1$. You should then get what you desire.