A geometry problem by Atul Kumar Ashish

We note that: $\begin{aligned} \tan (45-k)^\circ & = \dfrac {\tan 45^\circ - \tan k^\circ}{1+\tan 45^\circ \tan k^\circ} = \dfrac {1 - \tan k^\circ}{1 + \tan k^\circ} \end{aligned}$

Therefore, $\begin{aligned} \left(1+ \tan k^\circ\right)\left(1 + \tan (45-k)^\circ\right) & = \left(1+ \tan k^\circ\right)\left(1 + \dfrac {1 - \tan k^\circ}{1 + \tan k^\circ}\right) = 2 \end{aligned}$

Now, we have:

$\begin{aligned} P & = (1+\tan 1^\circ)(1+\tan 2^\circ)(1+\tan 3^\circ) \cdots (1+\tan 43^\circ)(1+\tan 44^\circ)(1+\color{#D61F06}{\tan 45^\circ}) \\ & = (1+\tan 1^\circ)(1+\tan 44^\circ)(1+\tan 2^\circ)(1+\tan 43^\circ) \cdots (1+\tan 22^\circ)(1+\tan 23^\circ)(1+\color{#D61F06}{1}) \\ & = \color{#3D99F6}{(1+\tan 1^\circ)(1+\tan (45-1)^\circ)}\color{#D61F06}{(1+\tan 2^\circ)(1+\tan (45-2)^\circ)} \cdots\color{#3D99F6}{(1+\tan 22)^\circ)(1+\tan (45-22^\circ)}\cdot \color{#D61F06}{2} \\ & = \underbrace{\color{#3D99F6}{2}\cdot \color{#D61F06}{2} \cdot \color{#3D99F6}{2} \cdot \color{#D61F06}{2} \cdots \color{#3D99F6}{2}}_{22 \text{ terms}}\cdot \color{#D61F06}{2} \\ & = 2^{23} \end{aligned}$

$\implies n = \boxed{23}$