I noticed that the graph of a binomial distribution resembled that of the Gaussian, so I scaled and translated the function to be symmetric about the $y$ axis. After some experimentation with scaling, I noticed that the function $$f(x)=\left(1+\frac{x}{\sqrt{n}}\right)^n\left(1-\frac{x}{\sqrt{n}}\right)^n$$ seemed to resemble the Gaussian function, and the graph seemed to approach that of the Gaussian for large values of $n$. I would like to know if $$e^{-x^2}=\lim_{n\to\infty}\left(1+\frac{x}{\sqrt{n}}\right)^n\left(1-\frac{x}{\sqrt{n}}\right)^n$$ for all values of $x$, and if this is a consequence of the central limit theorem (since the translations and $\sqrt{n}$ in the denominators suggest that the central limit theorem is somehow involved).
Edit: In the question Proving $\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n=\text{e}^x$., a similar result is proven, and it can be algebraically manipulated to prove my conjecture. However, I would like to know if there would be a probabilistic way to prove the identity that somehow involves the central limit theorem.
$$f(x)=\left(1+\frac{x}{\sqrt{n}}\right)^n\left(1-\frac{x}{\sqrt{n}}\right)^n=\left(1-\frac{x^2}{n}\right)^n$$ $$\log(f(x))=n\log\left(1-\frac{x^2}{n}\right)=-x^2-\frac{x^4}{2 n}-\frac{x^6}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$f(x)=e^{\log(f(x))}=e^{-x^2}\left(1-\frac{x^4}{2 n}+\frac{x^6 \left(3 x^2-8\right)}{12n^2}+O\left(\frac{1}{n^3}\right) \right)$$
Plot the function and its approximation for $-\sqrt n \leq x \leq \sqrt n$ : even for $n=2$, they almost exactly overlap.
Interesting is to look at the areas $$A_n=\int_{-\sqrt n}^{+\sqrt n} \left(1+\frac{x}{\sqrt{n}}\right)^n\left(1-\frac{x}{\sqrt{n}}\right)^n \,dx=\sqrt{\pi }\, n^{3/2}\, \frac{ \Gamma (n)}{\Gamma \left(n+\frac{3}{2}\right)}$$ $$A_n=\sqrt{\pi}\left( 1-\frac{3}{8 n}+\frac{25}{128 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ which is almost the same for the approximation and $$\int_{-\infty}^{+\infty}e^{-x^2}\,dx=\sqrt{\pi }$$