For all integers $n$, we define $\xi_n$ as follows: $\begin{cases} \xi_n = 1 & \text{if } n \equiv 0 \pmod{4} \text{ or } n \equiv 1 \pmod{4} \\ \xi_n= -1 & \text{if } n \equiv 2 \pmod{4} \text{ or } n \equiv 3 \pmod{4} \end{cases}$ For all $n \in \mathbb{Z^+}$, let $f(n)= \xi_0 \dbinom{n}{0} + \xi_1 \dbinom{n}{1} + \xi_2 \dbinom{n}{2} + \cdots + \xi_n \dbinom{n}{n}.$ Find $\left \lfloor 100 \left( \displaystyle \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!} \right) \right \rfloor$.

**Details and assumptions**

As an explicit example, since $4 \equiv 0 \pmod{4}$, $\xi_4= 1$, whereas $\xi_6 = -1$ since $6 \equiv 2 \pmod{4}$. Note that $\xi_0= \xi_1= 1$.

The floor function $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$. For example, $\lfloor 3.25 \rfloor = 3, \lfloor 4 \rfloor= 4, \lfloor \pi \rfloor = 3$.

You might use a scientific calculator for this problem.

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