Evaluating $\{1+\sum\limits_{\mu=1}^{\infty}(-1)^{\mu}\frac{z^{2\mu}}{2^{2^{\mu}}\cdot(\mu !)^2}\}^2$ Question
Calculate the following formula,
$$\{1+\sum\limits_{\mu=1}^{\infty}(-1)^{\mu}\frac{z^{2\mu}}{2^{2^{\mu}}\cdot(\mu !)^2}\}^2$$
where $z$ is arbitrary.
We are given the hint that we can quote the identity:
$$\sum\limits_{\mu=0}^{\nu}(\text{C}_{\nu}^{\mu})^2=\text{C}_{2\nu}^{\nu}=\frac{(2\nu)!}{(\nu!)^2}$$
Attempt
Below is my attempt:
\begin{align}
\{1+\sum\limits_{\mu=1}^{\infty}(-1)^{\mu}\frac{z^{2\mu}}{2^{2^{\mu}}\cdot(\mu !)^2}\}^2 &=\{\sum\limits_{\mu=0}^{\infty}(-1)^{\mu}\frac{z^{2\mu}}{2^{2^{\mu}}\cdot(\mu !)^2}\}^2\\ &=\sum\limits_{\mu=0}^{\infty}(-1)^{\mu}\frac{z^{2\mu}}{2^{2^{\mu}}\cdot(\mu !)^2}\sum\limits_{\nu=0}^{\infty}(-1)^{\nu}\frac{z^{2\nu}}{2^{2^{\nu}}\cdot(\nu !)^2}\\
&=\sum_{\mu+\nu=0}^{\infty}(-1)^{\mu+\nu}\frac{z^{2(\mu+\nu)}}{2^{2^{\mu}+2^{\nu}}\cdot(\nu!)^2 (\mu!)^2}
\end{align}
But I can't find a proper way to use the identity which is supported by the hint.\
Thanks for your help in advance!
 A: As I've said in a comment (now deleted), there's a typo. The copy I have poses the problem as $$\left\{1+\sum_{\mu=1}^\infty(-1)^\mu\frac{z^{2\mu}}{2^{2\mu}\cdot(\mu!)^2}\right\}^2\stackrel{?}{=}1+\sum_{\nu=1}^\infty(-1)^\nu\frac{(2\nu)!z^{2\nu}}{2^{2\nu}\cdot(\nu!)^4}$$ with the hint given by you (typeset mistakenly in my copy). This is what holds indeed:
\begin{align*}
\left\{\sum_{\mu=0}^\infty\frac{(-1)^\mu z^{2\mu}}{2^{2\mu}\mu!^2}\right\}^2
&=\sum_{\lambda=0}^\infty\sum_{\mu=0}^\infty\frac{(-1)^{\lambda+\mu}z^{2(\lambda+\mu)}}{2^{2(\lambda+\mu)}\lambda!^2\mu!^2}
\\\color{gray}{[\lambda+\mu=\nu]}
\quad&=\sum_{\nu=0}^\infty\sum_{\mu=0}^\nu\frac{(-1)^\nu z^{2\nu}}{2^{2\nu}\mu!^2(\nu-\mu)!^2}
\\&=\sum_{\nu=0}^\infty\frac{(-1)^\nu z^{2\nu}}{2^{2\nu}\nu!^2}\sum_{\mu=0}^\nu\left(\frac{\nu!}{\mu!(\nu-\mu)!}\right)^2
\\\color{gray}{[\text{using the hint}]}
\quad&=\sum_{\nu=0}^\infty\frac{(-1)^\nu z^{2\nu}}{2^{2\nu}\nu!^2}\frac{(2\nu)!}{\nu!^2}=\sum_{\nu=0}^\infty(-1)^\nu\frac{(2\nu)!z^{2\nu}}{2^{2\nu}\nu!^4}.
\end{align*}
A side note: the sum being squared is the Bessel function $J_0(z)$.
