Observation: $\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$ I have observed that this 
$$\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$$

(It can be fitted better than this, but I did not want them to completely overlap)
I want to learn how I would find $a_n$, or if this relationship is technically asymptotic to begin with.
 A: This is a nice observation!
The Taylor series for $$\operatorname{sinc} x := \frac{\sin x}{x}$$
at $x = 0$ is
$$\operatorname{sinc} x \sim 1 - \frac{1}{6} x^2 + \frac{1}{120} x^4 - \cdots + (-1)^{k / 2} \frac{1}{(2k + 1)!} + \cdots , $$
and so the series for $\operatorname{sinc}^n x$ is
$$\phantom{(\ast)} \qquad \operatorname{sinc}^n x \sim 1 - \frac{1}{6} n x^2 + \left( \frac{1}{72} n^2 - \frac{1}{180} n \right) x^4 - \cdots \qquad (\ast).$$
In principle it's not so hard to derive an expression for the coefficient of the $x^{2k}$ term, but its salient feature for our purposes is that its leading term (in $n$) is $$\frac{(-1)^{k / 2} n^k}{6^k k!}.$$
On the other hand, since $$\operatorname{sinc}^n 0 = \exp(-a_n \cdot 0^2) = 1 ,$$ we can naively attempt to align the two functions by choosing a parameter $a_n$ to match the Taylor series of the two functions to as high of an order as we can: The Taylor series of $\exp(-a_n x^2)$ is
$$\phantom{(\ast\ast)} \qquad \exp(-a_n x^2) \sim \sum_{k = 0}^{\infty} \frac{(-a_n x^2)^k}{k!} = 1 - a_n x^2 + \frac{1}{2} a_n^2 x^4 - \cdots + \frac{(-1)^{k / 2}}{k!} a_n^k x^{2k} + \cdots \qquad (\ast\ast),$$
so we can match the series $(\ast)$ and $(\ast\ast)$ to second order (in fact, third order) by taking $a_n$ so that their quadratic terms agree, that is, by setting $a_n = \frac{1}{6} n$. But then we see that 
$$\exp\left(-\frac{n}{6} x^2\right) \sim 1 - \frac{1}{6} n x^2 + \frac{1}{72} n^2 x^4 - \cdots + \frac{(-1)^{k / 2} n^k}{6^k k!} x^{2k} + \cdots.$$
But the coefficient of this general term is precisely the same as the leading term (in $n$) of the coefficient of the $x^{2k}$ term for the series for $\operatorname{sinc}^n x$, which we identified above.
With this in hand, we can make precise in various ways the notion that the two functions are asymptotic as $n \to \infty$. One way to make this more tractable is to normalize the widths of the functions and compare
$$\operatorname{sinc}^n \left( \frac{x}{\sqrt{n}}\right) \sim 1 - \frac{1}{6} x^2 + \left(\frac{1}{72} - \frac{1}{180 n} \right) x^4 - \cdots$$
with
$$\exp\left[-\frac{n}{6} \left( \frac{x}{\sqrt{n}}\right)^2\right] = \exp\left(-\frac{1}{6} x^2\right) \sim 1 - \frac{1}{6} x^2 + \frac{1}{72} x^4 - \cdots;$$ these differ by a series $$- \frac{1}{180 n} x^4 + \cdots ,$$ which in particular is $O(n^{-1})$.
A: You can understand this using probability theory.
The function $x\mapsto{\sin(x)\over x}$ is the characteristic function of a random variable $X$ that is uniformly distributed over $(-1,1)$. This random variable has mean $\mu=0$ and variance $\sigma^2=1/3$. By the central limit theorem, the normalized sum ${1\over \sqrt{n}}\sum_{i=1}^n X_i$ has a Gaussian limit, so 
$$\left[{\sin(x/\sqrt{n})\over (x/\sqrt{n})}\right]^n\to \exp(-x^2\sigma^2/2), $$
or
$$\left[{\sin(x)\over x}\right]^n\approx \exp(-n x^2/6)$$
for small $x$. 
