Finding value of $r$ for which $n^r(1-X_{(n)})$ converges in distribution

Let $$X_1,X_2,\ldots,X_n$$ be i.i.d and $$X_{(n)}=\max_{1 \leq i \leq n}X_i$$. If $$X_i \sim \text{Beta}(1,\beta)$$, find a value of $$r$$ such that $$n^r(1-X_{(n)})$$ converges in distribution.

So, I tried proving convergence in probability and then using a specific $$\epsilon$$ but I got stuck.

Hint: write down an expression for $$P(n^r ( 1 - X_{(n)}) \ge t)$$, and choose $$r$$ so that the expression converges to some limit as $$n \to \infty$$.
$$P(n^r(1-X_{(n)}) \ge t) = P(X_{(n)} \le 1 - t n^{-r}) = P(X_1 \le 1 - tn^{-r})^n = ( 1 - (tn^{-r})^\beta)^n$$ where the last step uses the CDF of the $$\text{Beta}(1, \beta)$$ distribution. To choose the appropriate value of $$r$$, it may help to recall the limit $$(1 + \frac{u}{n})^n \to e^u$$ for real $$u$$.
• I think I have the answer r=1/$\beta$, right? – Bayesian guy Sep 1 '19 at 2:07