Finding the residues of poles Consider the equation $\mathcal{F}(\lambda)=0\ \ \ \forall\ \lambda = \lambda_{n},\ n \in \mathbb{N}$.
I understand that the expression $\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}$ has poles of order 1 exactly at $\lambda_{n}$ because $\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'(\lambda)}}{(\lambda- \lambda_{1})...(\lambda-\lambda_{n})}$.
I wonder how I might expand the expression $\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}$ about $\lambda_{n}$ to find out that the residue of the poles is 1.
Any ideas?
Edit to the post with an alternative answer:
I've found out that, expanding about $\lambda = \lambda_{n}$, with $\mathcal{F'}(\lambda_{n})\neq0$, we obtain
$\frac{\mathcal{F'}(\lambda)}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})}{\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})} = \frac{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...}{(\lambda-\lambda_{n})\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})^{2}\mathcal{F''}(\lambda_{n})+...} = \frac{1}{\lambda - \lambda_{n}}+...$,
so that the residue at all eigenvalues is 1.
I took this evaluation out from a paper, so I am not really sure about a couple of things I have written -  have we used the Taylor series to expand each of $\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})$ and $\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})$ about $\lambda=\lambda_{n}$? I know this is a silly question, as you've already used the Weierstrass factorisation theorem, but I still want to clarify things up.
 A: Your last approach is correct. Here is a slight variation of the argument that avoids Taylor expanding. If $\mathcal{F}$ is analytic at $\lambda$ and $\mathcal{F}(\lambda) = 0$ with $\mathcal{F}'(\lambda) \not= 0$ then we can write $\mathcal{F}(z) = (z-\lambda)g(z)$ where $g(z) = \frac{\mathcal{F}(z)}{z-\lambda}$. Now since $\lim\limits_{z\to\lambda}g(z) = \mathcal{F}'(\lambda)$ we have that $g$ is analytic at $z=\lambda$ (it has a removable singularity) and
$$\frac{\mathcal{F}'(z)}{\mathcal{F}(z)} = \frac{1}{z-\lambda} + \frac{g'(z)}{g(z)}$$
Since $\mathcal{F}'(\lambda) \not= 0$ we have $g(\lambda)\not=0$ so the last term is analytic at $z=\lambda$ and the residue can be read off from the first term.

As a sidenote to your first attempt: when $\mathcal{F}$ is entire and has zeros for all $z\in\{\lambda_n\}_{n=1}^\infty$ you cannot simply write $\mathcal{F}(z) = A(z-\lambda_1)(z-\lambda_2)\cdots$ as we can do when $\mathcal{F}$ is a polynomial. However there is a very nice theorem called Weierstrass factorization theorem that gives us the functional form of $\mathcal{F}$. It is slightly more complicated:
$$\mathcal{F}(z)=z^m e^{h(z)} \prod_{n=1}^\infty \left(1 - \frac{z}{\lambda_n}\right)E_{p_n}\!\!\left(\frac{z}{\lambda_n}\right)$$
where $h$ is some analytic function, $m$ is the order of the zero at $z=0$, $p_n$ is some set of integers and
$$E_{n} = \left\{\matrix{1 & n=0\\\exp(z + \frac{z^2}{2} + \ldots + \frac{z^n}{n}) & n> 0 }\right.$$
If you want you can also use this formula to solve your problem. By taking the logarithmic-derivative we get
$$\frac{\mathcal{F}'(z)}{\mathcal{F}(z)} = \frac{m}{z} + \sum_{n=1}^\infty \frac{1}{z-\lambda_n} + \text{(analytic function)}$$
so the residue $\text{Res}\left[\frac{\mathcal{F}'(z)}{\mathcal{F}(z)};\lambda\right]$ is equal to the order of the zero of $\mathcal{F}$ at $z=\lambda$ which in your case is just $1$ since all zeros are simple.
