statistics – Exact confidence interval for the uniform parameter

I am given the following.

$$ X_1, X_2, …, X_n sim U (0, theta) $$
and I want to get an exact confidence interval of $ theta $ without using the normal approximation.

What I know up to now is that $ Y_n = max (x_1, x_2, …, x_n) $ is the MLE for $ theta $ and $ frac {n + 1} {n} Y_n $ is the unbiased estimator.

I was asked to find the confidence interval based on this estimator and that's what I tried.

Let $ W = frac {n + 1} {n} Y_n / theta $ then

$$ Pr[l
$$ Pr[W
$$ Pr[W
or $ l $ and $ u $ represent the lower and upper bound of the confidence interval.

Now, here is what troubles me.

When I solve for the confidence interval, I get
$$ frac {Y_n} {^ n sqrt {1- alpha / 2}} < theta < frac {Y_n} {^ n sqrt { alpha / 2}} $$
and algebraically the $ frac {n + 1} {n} $ has become irrelevant.

even though I have $ l = frac {n + 1} {n} ^ n sqrt { alpha / 2} $ and $ u = frac {n + 1} {n} ^ n sqrt {1- alpha / 2} $

In trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $ frac {n + 1} {n} Y_n $ is greater than the MLE, $ Y_n $how is it that the confidence interval is not centered on this unbiased estimator?

I know that this question is very strange and that I could not find something similar to my argument online, I would really appreciate your help.