$(1+\tan 1^o)(1+\tan 2^o)...(1+\tan 45^o)=2^{25-n}$ then n must be-

The LHS can be rewritten as : 
$\left\{(1+\tan 1^o)(1+\tan 44^o)\right\}\left\{1+\tan 2^o)(1+\tan 43^o\right\}.....\left\{1+\tan 45^o\right\}$

Now, 
$\tan (1^0+44^0)=\frac{\tan 1+\tan 44}{1-\tan 1\tan 44}$
$\Rightarrow 1=\frac{\tan 1+\tan 44}{1-\tan 1\tan 44}$
$\Rightarrow \tan 1+\tan 44+\tan 1\tan 44=1$
$\Rightarrow 1+\tan 1+\tan 44+\tan 1\tan 44=2$
$\Rightarrow (1+\tan 1)(1+\tan 44)=2$
Similarly,
$(1+\tan 2)(1+\tan 43)=(1+\tan 3)(1+\tan 42)=...=(1+\tan 22)(1+\tan 23)=2$
$\Rightarrow 2^{23}=2^{25-n}$
$\Rightarrow n=2$