Consider the following finite power tower:
$$\Large \tan(1°)^{\tan(2°)^{\tan(3°)^{\cdot^{\cdot^{\cdot^{\tan(44°)^{\tan(45°)}}}}}}}$$
I'm wondering if there is a way to solve this that doesn't rely on brute force (i.e. simply typing the entire power tower out). As, I see it, there are two possibilities:
- There's some magic trig identity that simplifies this to a small closed form. I haven't found any such yet.
- I need to develop some sort of approximation scheme for a computer.
Is (1) possible? If not, how do I do (2)?
Edit: For clarification, I have manually typed out the problem inside Desmos and arrived at the following answer: 
I'd just like to know if this answer is correct, and if so, how I would go about finding an easier way to achieve this answer (whether that be by using a trig identity or some form of computerization, I don't mind) without having the type out every step like I've done above. Thank you,
Taking them five at the time, the results are $$\left( \begin{array}{ccc} \text{from} & \text{to} & \text{value} \\ 1 & 5 & 0.0535625370657532607362393 \\ 1 & 10 & 0.4575153711252385756282372 \\ 1 & 15 & 0.2482699982569751321101037 \\ 1 & 20 & 0.2797539387187735772940906 \\ 1 & 25 & 0.2773295827116058036618656 \\ 1 & 30 & 0.2774030591794901804478588 \\ 1 & 35 & 0.2774023124430804526117091 \\ 1 & 40 & 0.2774023141041003428422647 \\ 1 & 45 & 0.2774023141038327352949884 \end{array} \right)$$
For the fun of it, with an absolute error of $5.79\times 10^{-21}$, the final value is almost $$\Bigg[e^{-42-\frac{34}{e}-47 e+\frac{7}{\pi }+30 \pi }\,\, \pi ^{23 e-15} \,\,\tan ^{31}(e \pi )\,\, \sec ^{13}(e \pi ) \Bigg]^{\frac 1 {15}}$$ given by a friend who enjoys this kind of problems.