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Relevant wiki: Gamma Function

Since $\Gamma{\left(\dfrac{1}{2}\right)} = \sqrt{\pi}$ we have that

$\Gamma{\left(\dfrac{3}{2}\right)} = \Gamma{\left(1+\dfrac{1}{2}\right)} = \dfrac{1}{2}\Gamma{\left(\dfrac{1}{2}\right)} = \dfrac{1}{2}\sqrt{\pi} \\ \Gamma{\left(\dfrac{5}{2}\right)} = \Gamma{\left(1+\dfrac{3}{2}\right)} = \dfrac{3}{2}\Gamma{\left(\dfrac{3}{2}\right)} = \dfrac{3 \times 1}{2 \times 2}\sqrt{\pi} \\ \Gamma{\left(\dfrac{7}{2}\right)} = \Gamma{\left(1+\dfrac{5}{2}\right)} = \dfrac{5}{2}\Gamma{\left(\dfrac{5}{2}\right)} = \dfrac{5 \times 3 \times 1}{2 \times 2 \times 2}\sqrt{\pi} \\ \vdots \\ \Gamma{\left(\dfrac{2n+1}{2}\right)} = \Gamma{\left(1+\dfrac{2n-1}{2}\right)} = \dfrac{2n-1}{2}\Gamma{\left(\dfrac{2n-1}{2}\right)} = \dfrac{(2n-1) \times (2n-3) \times \cdots \times 3 \times 1}{\underbrace{2 \times 2 \times \cdots \times 2 \times 2}_{n \space \space 2's}}\sqrt{\pi} \\ \implies \dfrac{2n-1}{2}\Gamma{\left(\dfrac{2n-1}{2}\right)} = \dfrac{(2n-1)!!}{2^n}\sqrt{\pi} \\ \therefore (2n-1)!! = \dfrac{2^{n-1}(2n-1)}{\sqrt{\pi}}\Gamma{\left(\dfrac{2n-1}{2}\right)} \\ \text{So } \ (2n-1)!! = 9999!! \implies 2n-1 = 9999 \\ \therefore n = \ \boxed{5000}$ .

Lemma:

$\dfrac{2^{n-1} (2n-1)}{\sqrt{\pi}} \Gamma \left( \dfrac{2n-1}{2} \right) = (2n-1)!!$

Proof:

Using the property $\Gamma(s+1) = s \Gamma(s)$ , we can write

$\begin{aligned} \Gamma \left( \dfrac{2n-1}{2} \right) &= \left( \dfrac{2n-3}{2} \right) \Gamma \left( \dfrac{2n-3}{2} \right) & \\ &= \left( \dfrac{2n-3}{2} \right) \left( \dfrac{2n-5}{2} \right) \Gamma \left( \dfrac{2n-5}{2} \right) & \\ & \qquad \qquad \qquad \vdots & \\ &= \dfrac{(2n-3)!!}{2^{n-1}} \Gamma \left( \dfrac 12 \right) & \\ &= \dfrac{(2n-3)!!}{2^{n-1}} \sqrt{\pi} & \end{aligned}$

Thus

$\begin{aligned} & \dfrac{2^{n-1} (2n-1)}{\sqrt{\pi}} \Gamma \left( \dfrac{2n-1}{2} \right) = \dfrac{2^{n-1} (2n-1)}{\sqrt{\pi}} \cdot \dfrac{(2n-3)!!}{2^{n-1}} \sqrt{\pi} = (2n-1)!! & & \\ & & & \mathbf{Q.E.D.} \end{aligned}$

So for our problem

$(2n-1)!! = 9999!! \implies n = \boxed{5000}$