[Revised]Proving the expected \bold{density} of being the Nth order statistics is decreasing in sample size (Sorry that I've previously formulated the question in a wrong way, which confused everyone including myself. This is a better version of the question. Thanks!)
Here's another order statistics question that I wish to ask.
Question
Consider $n$ random variables $x_1, x_2,\cdots x_n\overset{iid}{\sim} D$. Where $D$ is some unimodal on 0, symmetric, continuous distribution with a finite variance (P.S. This condition might be overly restrictive, suggestions on loosening it would be greatly appreciated! ). The PDF for nth order statistics is 
$nF_{D}(x)^{n-1}f(x).$
I'm interested in the properties of the following "expected density" (I'm not sure there's a better way to put it) of the nth order PDF:
$\displaystyle\int_{-\infty}^{+\infty}(n-1)F_{D}(x)^{n-2}f_{D}(x)\times f_{D}(x)\:dx$, 
which simplifies to 
$\displaystyle\int_{-\infty}^{+\infty}(n-1)F_{D}(x)^{n-2}f_{D}(x)^2\:dx$
What I want to show is that this expression decreases as $n$ increases.
What I have gotten so far:
By an integration by parts trick, we can show that the above can be expressed as:
$\int_{-\infty}^{\infty}f_{D}(x)dF_{D}(x)^{n-1}$
$=f_{D}(x)F_{D}(x)^{n-1}|_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}F_{D}(x)^{n-1}df_{D}(x)$
$=-\int_{-\infty}^{+\infty}F_{D}(x)^{n-1}f^{\prime}_{D}(x)dx.$
Intuitively, I can see that as $-f^{\prime}_{D}(x)$ is positive on $x>0$ and negative on $x<0$. And as $n$ increases $F^n_{D}(.)$ shifts more mass to extreme values where $f^{\prime}_{D}(x)$ is very close to 0. So, eventually the whole integral will become smaller as $n$ increases. But I'm not sure how to proceed this argument formally. Any help will be greatly appreciated!
 A: Expectation of a conditional probability is the total probability, thus the expected probability would be just the probability that $x_k$ is the maximum, which in the case of continuous distribution is $1/n$, from which decreasing in $n$ immediately follows. (This was pointed out in the comments by whuber). In this answer, I also show how to formulate the expectation integral correctly and evaluate it to be $1/n$ - this is not needed for proving the result, but may be useful for learning purposes to see what went wrong in the original approach.
The first integral is incorrect. The conditional probability of $x_k$ being the maximum (conditional on  of $x_k$) is simply the probability that all others are less than (or equal, which does not matter in the continuous case) $x_k$: 
\begin{equation}
\mathbb{P}(x_k = \max(x_1,\ldots,x_n) \mid x_k) = \mathbb{P}(x_1\leq x_k,\ldots,x_{k-1}\leq x_k,x_{k+1}\leq x_k,\ldots,x_n \leq x_k)
\end{equation}
Applying the independence property and the definition of the CDF $F_D$, we obtain
\begin{equation}
=\mathbb{P}(x_1\leq x_k)\,\ldots\,\mathbb{P}(x_{k-1}\leq x_k)\,\mathbb{P}(x_{k+1}\leq x_k)\ldots\,\mathbb{P}(x_n\leq x_k)=F_D(x_k)^{n-1}.
\end{equation}
Then, to obtain the expectation of this random variable, we average over different values of $x_k$ weighting by the PDF of $x_k$:
\begin{equation}
\mathbb{E}(\mathbb{P}(x_k = \max(x_1,\ldots,x_n) \mid x_k)) = \mathbb{E}(F_D(x_k)^{n-1}) = \int_{-\infty}^{\infty}f_D(x)F_D(x)^{n-1}dx.
\end{equation}
Using the chain rule, the derivative of $\frac{1}{n}F_D(x)^n$ w.r.t. $x$ is $\frac{1}{n}f_D(x)\,n\,F_D(x)^{n-1}$ $= f_D(x)F_D(x)^{n-1}$, and thus our last integral equals 
\begin{equation}
= \left. \frac{1}{n}F_D(x)^n \right|_{x=-\infty}^{x=\infty} = \frac{1}{n}.
\end{equation}
