The examples given with base 10 and 2 in the question are positional bases. In a positional base 1, you only got one digit, with no value: 0. All positions will have zero value, and you can only represent one number: 0. – Bijective base 1 would be one way to make it funcitonal, but that isn't a positional base.
Generically you don't know without examing the presumed "basis" vectors. You do know that three vectors are sufficient (x,y,z) to span 3-space; any fourth vector must be a linear combination of (x,y,z). There is no more room.
The reason why $1^\infty$ is indeterminate, is because what it really means intuitively is an approximation of the type $ (\sim 1)^ {\rm large \, number}$. And while $1$ to a large power is 1, a number very close to 1 to a large power can be anything.....
The polynomial $x^n-1$ is interesting in for example many signal processing applications. It is easy to find special cases for factorizations, for example the famous conjugate rule: $$x^2-1 = (x+1) (x-1)$$
Pretend K=3 That means (K+1)= 4 This means you'd be dividing 3*2*1 by 4*3*2*1. Consider how you'd cancel out multiples by dividing them. Like how (2 (5+x))/2 would just equal 5+x. Following that idea we'd pretty much be able to cancel out every number in the numerator, so long as its also in the denominator. This would end up canceling every number except for 4 which equals (K+1). Essentially ...
You don't really need a formal induction here: the formula is equivalent to $$ (1-a) (1+a+a^2+\dots+a^ {n-1})=1-a^n, $$ a high-school factorisation formula, that you can prove doing the multiplication in the l.h.s.: $$\begin {alignedat} {6}1&+ {}&a&+ {}&a^2&+ {}&\dotsm\dotsm&+ {}&a^ {n-1} \\ &- {}&a&- {}&a^2&-&a^3-\dotsm&- {}&a^ {n-1}&-a^ {n ...
