$\dfrac{2}{\pi} = \dfrac{\sqrt 2}{2} \cdot \dfrac{\sqrt {2+\sqrt 2}}{2} \cdot\dfrac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdots $ One can  show inductively that
$$
\cos \frac{\pi}{2^{n+1}}\ = \frac{\sqrt {2+\sqrt {2+\sqrt {2+\sqrt {\cdots+\sqrt {2 }}}}}}{2},
$$
with $n$ square roots in the right side of the equation.
The second part of the question was to deduct the following from the first part:
$$\frac{2}{\pi} = \frac{\sqrt 2}{2} \cdot  \frac{\sqrt {2+\sqrt 2}}{2} \cdot\frac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdot \cdots $$
with the hint to use the following limit:
$$\lim_{n\to \infty}\cos\Big(\frac{t}{2}\Big)\cos\Big(\frac{t}{2^2}\Big)\cdots\cos\Big(\frac{t}{2^n}\Big) = \frac{\sin t}{t}.$$
A hint or some general intuition will be appreciated.

Using the trogonometric identity
$$
\sin (2a)=2\sin a\,\cos a\qquad\text{or}\qquad \cos a=\frac{\sin 2a}{2\sin a},
$$
provided that $\,\sin a\ne 0,\,$ we obtain that
$$
\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\cdots\frac{\sin (x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin x}{2^n\sin(x/2^n)}.
$$
Hence
$$
\lim_{n\to\infty}\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{x},
$$
since
$$
\lim_{t\to 0}\frac{\sin(tx)}{t}=x.
$$
In particular,
$$
\prod_{n=1}^\infty \cos\left(\frac{\pi}{2^{n+1}}\right)=\frac{2}{\pi}.
$$
