Prove that $\operatorname{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\sigma^2dt$ with $\mathbb{E}[X]:=p(u-d)+d$

Question

I have to prove that $\mathbb{E}[X^2]-\mathbb{E}[X]^2=\sigma^2dt$ for $\mathbb{E}[X]:=p(u-d)+d$ as from title, and with:

• $p:=\frac{e^{rdt}-d}{u-d}$;

• $u:=e^{\sigma\sqrt{dt}}$;

• $d:=e^{-\sigma\sqrt{dt}}$.

I arrive to say that $e^{r\sigma dt\sqrt{dt}}+e^{-r \sigma dt \sqrt{dt}}-1-e^{2rdt}$ but now I'm stuck. How can I continue?

Thanks in advance for any help.