Tan-Tadan

$(1+tan(1^{\circ}))(1+tan(2^{\circ}))...(1+tan(45^{\circ}))$

$(1+tan(1^{\circ}))(1+tan(2^{\circ})) .... ( 1 + tan(45 - 1)^{\circ} \times 2$

$(1+tan(1^{\circ}))(1+tan(2^{\circ})) .... ( 1 + \dfrac{ 1 - tan1^{\circ}}{1 + tan1^{\circ}})\times 2$

$(1+tan(1^{\circ}))(1+tan(2^{\circ})) ....(\dfrac{2}{1+tan1^{\circ}})\times 2$

similarly for 2 and 43 , 3 and 42.......... so we can see 22 pairs , so they will give $2^{22}$ and other 2 due to 1 + tan45 thus n=23

Note that $\left(1+\tan\alpha^\circ\right)\times{\left(1+\tan(45-\alpha)^\circ\right)}=2$ .

Now, grouping the expression in pairs like below:

$p_{1}=\left(1+\tan1^\circ\right)\times{\left(1+\tan44^\circ\right)}=2\\p_{2}=\left(1+\tan2^\circ\right)\times{\left(1+\tan43^\circ\right)}=2\\\vdots\\p_{22}=\left(1+\tan22^\circ\right)\times{\left(1+\tan23^\circ\right)}=2\\p_{23}=\left(1+\tan45^\circ\right)=2$ .

Hence, $p_{1}\times{p_{2}}\times{\ldots}\times{p_{23}}=\underbrace{2\times{2}\times{\ldots}\times{2}}_\text{23 times}=2^{23}$