Limit of n times Beta(1,n) variables when n goes to infinity Wikipedia states the following:
(https://en.wikipedia.org/wiki/Beta_distribution#Special_and_limiting_cases)
$$\lim_{n \to \infty} nB(1,n) = \operatorname{Exponential}(1).$$
I'm having trouble deriving this result. This is how far I get:
$$B(1,n) = \frac{1}{\operatorname{Beta}(1,n)} (1-x)^{n-1} = \frac{\Gamma(1+n)}{\Gamma(1)\Gamma(n)} (1-x)^{n-1} = n(1-x)^{n-1} $$
I think I'm supposed to make use of the fact that $\lim_{n \to \infty} (1-\frac{x}{n})^{n} = e^{-x}$, but I don't see how.
 A: First, let's get a sense why this should be true.  The density of a Beta$(1,n)$ variable which has been multiplied by $n$ should be proportional to
$$\left(\frac{x}{n}\right)^{1-1}\left(1 - \frac{x}{n}\right)^{n-1} \propto \left(1 - \frac{x}{n}\right)^{n-1} \approx e^{-x}$$
(for large $n$, anyway), so it surely looks exponential.

A rigorous and relatively elementary way to demonstrate the result is to work with the distribution function (CDF) $F_n$.  That is, suppose $Y_n$ has a Beta$(1,n)$ distribution (for $n \gt 0$) and let $X_n=nY_n$.  By definition,
$$F_n(x) = \Pr(X_n \le x) = \Pr(nY_n \le x) = \Pr\left(Y_n \le \frac{x}{n}\right).$$
When $0 \lt x \lt n$, this probability is given by the Beta integral, proportional to
$$\Pr\left(Y_n \le \frac{x}{n}\right) \propto \int_0^{x/n} (1-y)^{n-1} dy = -\frac{1}{n} (1-y)^n|_0^{x/n} \propto 1 - \left(1 - \frac{x}{n}\right)^n.\tag{*}$$
When $x \ge n$, this probability is $1$ (it's no longer given by the integral).
(Notice how freely we may drop any multiplicative constants, like that factor of $1/n$, that do not depend on $y$ or $x$: in the end we only need to establish that the limiting function rises from $0$ to a finite value in the limit as $x\to\infty$.  The function then can be divided by that limiting value to produce a genuine distribution.)
The right hand side is often used to define the exponential in the sense that
$$e^{-x} = \lim_{n\to \infty} \left(1 - \frac{x}{n}\right)^n.$$
Since for any $x \gt 0$ and $n\to \infty$ eventually $x\lt n$, the limiting value of $F_n(x)$ is the limiting value of $(*)$: we don't have to worry about the fact that $F_n(x)=1$ when $n$ is small.  Furthermore, for $x\le 0$, $F_n(x)=0$ always and so its limiting value obviously is $0$ in such cases.

To illustrate the analysis, this figure plots $F_1$ (blue), $F_2$ (red), $F_4$ (gold), $F_8$ (green), and the limiting distribution (dashed, in gray).  Evidently the distributions $F_n$ converge down to their limiting value everywhere $x \gt 0$.
The normalizing constant turns out to be unity, because the limiting value of $1 - e^{-x}$ as $x\to\infty$ is $1$: it already is a valid distribution function.
This shows that $F_n(x)$ approaches $F(x)=1 - e^{-x}$ arbitrarily closely for any $x\gt 0$ for sufficiently large $n$ and otherwise is $0$.  This is the standard Exponential distribution.  Therefore whenever $(Y_n)$ is a sequence of random variables with Beta$(1,n)$ distributions, the distributions of the random variables $(nY_n)$ converge to the standard Exponential distribution, QED.

Addendum
It might be worthwhile to show what can go wrong with an analysis of densities (PDFs).
For any $n=1,2,\ldots,$ define a "uniform $n$-distribution" to be an equally-weighted mixture of Normal distributions, each with standard deviation $2^{-2n}$ and located at the odd multiples of $2^{-n}$: that is, at $k2^{-n}$ for $k=1, 3, \ldots, 2^n-1$.  Here is a plot of densities of the uniform $n$-distributions for $n=1,2,3,4$:

The uniform $n$-distribution has $2^{n-1}$ spikes and those spikes are occupying exponentially narrower portions of the gaps between their peaks.  Because this is a finite mixture of Normal distributions it has a very nice density which is bounded, nonzero, and infinitely differentiable everywhere--one could scarcely complain of any mathematical "pathology."  However, this family has been constructed to ensure that the limit of these densities is almost everywhere zero.  (This is not hard to prove, but the details might be distracting here, so I will rely on the figure to make the point.)  Note that zero itself is a nice function, too: bounded and infinitely differentiable.  It's just impossible to normalize it to unit area!
Nevertheless, this sequence of distribution functions does have a limiting distribution function: it is the (usual) Uniform$(0,1)$ distribution.  Here is a picture corresponding to the previous one, showing their distribution functions in the same colors:

The limit is a uniform distribution because between any $0 \le a \lt b \le 1$ there are approximately $(b-a)2^{n-1}$ spikes, each with almost all its probability (totaling $2^{1-n}$) concentrated between $a$ and $b$, for a total probability close to $b-a$: that is nearly uniform.  In the picture you see a sequence of staircase-like graphs with smaller and smaller steps squeezing down to the slanted ramp (gray dots): that's the Uniform$(0,1)$ CDF.
The problem isn't restricted to densities that converge to $0$.  Pick $0 \lt p\lt 1$ and let $(X_n)$ be a sequence of random variables with distributions that are a mixture of $p$ times any absolutely continuous distribution $F$ you want and $1-p$ times the uniform $n$-distributions.  This sequence of densities converges to $p$ times the density of $F$.  Although that is a nonzero function, it's not a valid density because it integrates to $p$, not to $1$.
The moral is that even when a sequence of very nice density functions converges, it doesn't necessarily converge to a density function.
A: Thanks to the comments I can now formulate the answer myself, I think:
We know that Y=nX, so we can use a general formula for the pdf of a transformed variable (link of Christoph Hanck): 
$\rho_Y(y)=\frac{\rho_X(x)}{|f'(x)|}$ 
$f'(x)=\frac{d(nx)}{dx}=n$
$\rho_Y(y) = \frac{n(1-y/n)^{n-1}}{n}$
$\lim_{n \to \infty} (1-y/n)^{n-1}=e^{y}$
