Self multiplication of a CDF degenerates into a Dirac Delta? Because of some work I need to do, I bumped into this problem. A CDF $F(x):\mathbb{R} \mapsto [0,1]$ of some random variable is raised to the n-th power. The function is continuous all over its domain and has the typical characteristics of CDFs (monotonic ascending etc.). What happens when I consider the following?
$$\lim_{n\to+\infty}F^n(x)$$
I plotted some stuff and as the power increases, I get an interesting situation. In some cases, it seems like the function tends to a dirac delta, but in other circumstances, I get different evidence. For example, using the Gaussian distribution I have the following:

Looks like the function approaches to a delta. In the picture this is not evident sorry, but as $n$ increases, the function looks like converging to a vertical line on the right. If I use an exponent like $n=10000$ I get a diagram with a rect vertical line on the right.
But if I use the exponential CDF:

Here no matter the exponent, The curve seems translating to the right.
So my question is are there conditions for which the following:
$$\lim_{n\to+\infty}F^n(x) = \delta_t(x)$$
holds? Where $t$ is the delta translation coefficient.

As $F$ is a CDF, we have $0 \le F(x) \le 1$ for all $x \in \def\R{\mathbb R}\R$. As $q^n \to 0$ exactly for $\lvert q \rvert  < 1$, we have 
$$ 
  F^n(x) \to \begin{cases} 1 & F(x) = 1\\ 0 & \text{otherwise} \end{cases}
$$
So, $F^n$ tends pointwise to $\chi_{\{F=1\}}$, the indicator function of the set, where $F$ equals $1$ and this is never equal to a dirac distribution. In both your examples, $F$ is never 1, so $F^n \to 0$ for the gaussian and the exponential CDFs.
