We want to calculate an expression for:
(v)^{n}
where:
 n is an integer.
 v is a vector.
First we need to understand what type of multiplication we are using, for instance does v²= v•v or v×v ?
In order to keep this general I will use the geometric product '*' (see Clifford algebra) which is defined as:
a*b = a•b + a×b
but we will assume the vectors anticommute which means:
v×v=0
So we can use the dot product.
If the vectors anticommute (as they do in euclidean space) then we will see that the result is not necessarily closed within the vector space, if 'n' is even then the result will be a pure scalar value and if 'n' is odd then the result will be a vector.
Attempt to use binomial theorem
When we calculated the powers of complex numbers we used the binomial theorem:
(a + i b)^{n}= 


(i)^{k} a ^{nk} b^{k} 
However there are a couple of differences in this case: these are more terms (so we would have to use a 'n'nomial instead of a binomial) but more importantly the terms don't all commute so we must use all permutations rather than integer multiplications of combinations which the binomial theorem gives.
see combinatorics
Square
So lets try the simplest case where: (v)² since a vector squares to a scalar value we get:
v² = v•v = 

v_{k}² 
Other powers
So even powers will have scalar values, here are the first few powers:
power  value  type 

v²  v•v  scalar 
v³  v•v*v  vector 
v^{4}  (v•v)*(v•v)  scalar 
v^{5}  (v•v)*(v•v)*v  vector 
So the even powers of v^{n} are ∏v_{k}
The odd powers are v^{n} are v*∏v_{(k1)}