sir, at 5:07, in line 22, if n is odd then it should be " smaller_ans(a,(n-1)/2)a " instead of ’ smaller_ansa ’
One syntax doubt
hello @chandreshmaurya
no , we already have computed small = fast_pow(a,n/2) i.e a^((n-1)/2)
then we take its sqaure i.e small= small * small.
so now small is now eqaul to a^(n-1)
to get a^n we again need to multiply a into a^(n-1) .
that is why in line 22 we are doing a*small
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