Hadamard Product of $e^z-1$

Assuming the formulas $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}\quad\text{and}\quad\frac{\sin(\pi z)}{\pi z}=\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right),$$ I want to find the Hadamard product of $e^z-1$. I think I've done the work correctly, but I end up having an extra factor of $2\pi i$ at the end (this is from Stein and Shakarchi's book and they list the answer for this particular problem). So, my work is \begin{align} e^z-1&=e^{z/2}\left(e^{z/2}-e^{-z/2}\right)\\ \\ &=e^{z/2}\left(e^{i(-iz/2)}-e^{-i(-iz/2)}\right)\\ \\ &=2ie^{z/2}\left(\frac{e^{i(-iz/2)}-e^{-i(-iz/2)}}{2i}\right)\\ \\ &=2ie^{z/2}\sin(-iz/2)\\ \\ &=2\pi i ze^{z/2}\frac{\sin(-iz/2)}{\pi z}\\ \\ &=2\pi i ze^{z/2}\frac{\sin\left(\pi(\frac{-iz}{2\pi})\right)}{\pi z}\tag{$\ast$}\\ \\ &=2\pi i ze^{z/2}\prod_{n=1}^\infty\left(1-\frac{\left(\frac{-iz}{2\pi}\right)^2}{n^2}\right)\\ \\ &=2\pi i ze^{z/2}\prod_{n=1}^\infty\left(1+\frac{z^2}{4n^2\pi^2}\right). \end{align}

The answer according to the book is $$e^z-1=e^{z/2}z\prod_{n=1}^\infty\left(1+\frac{z^2}{4n^2\pi^2}\right).$$Have I made a simple mistake in calculation, or are the two somehow compatible with each other?

Update: Thomas Andrews has pointed out a mistake in my infinite product formula. I've corrected that mistake.

$(\ast)$ The error occurs from this line to the next. Instead of multiplying and dividing by $z$, I should have multiplied and divided by $\frac{-iz}{2\pi}$. That would cause the appropriate cancellation.

Solution 1:

You've got the wrong product formula for $\frac{\sin \pi z}{\pi}$. According to this page: $$\frac{\sin\pi z}{\pi z}=\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2}\right)$$

Note the $z$ in the denominator.

You have:

$$e^z-1 =2ie^{z/2}\sin(-iz/2)$$

Letting $w=\frac{-iz}{2\pi}$ this is $\sin \pi w$ and we can substitute into:

$$\sin \pi w = \pi w\prod\left(1-\frac{w^2}{n^2}\right)$$

But notice $\pi w = -iz/2$. So this yields:

$$e^z -1 = 2i e^{z/2} (-iz/2) \prod_n \left(1+\frac{z^2}{4\pi^2 n^2}\right)\\= e^{z/2} z \prod_n \left(1+\frac{z^2}{4\pi^2 n^2}\right)$$