Let R be the collection of complex linear combinations of elements of G. The R is a finite dimensional vector space over the field of complex numbers spanned by the elements of G; thus there are elements g1,..., gr of G which form a basis of R.
Now suppose r is an element of R such that the traces Trace(gi . r) = 0 all vanish. Then we obtain Trace(rn) = 0 for all positive integers n by expressing rn - 1 as a linear combination of the gi. But then these identities imply that r = 0. Thus an element g of G is uniquely determined once we know Trace(gi . g) for all i (if Trace(gi . g) = Trace(gi . h) then apply the above argument to r = g - h).
Now we are given that each element of G satisfies gN = e. Thus the trace of any element of G is a sum of n numbers of the form exp(2 . k/N) for k = 1,..., N. But there are only finitely such sums. Thus by the previous paragraph there are only finitely many elements in G. (Exercise: use the above argument to provide an explicit bound).
We now show how to reduce the General Burnside Problem to the Ordinary Burnside Problem in this case. Let K be the field generated (over the field Q of rational numbers) by the matrix coefficients of the finite collection of generators of G. Let L be the subfield of K consisting of all algebraic numbers (elements satisfying a polynomial with rational coefficients). Since K is finitely generated L is a finite extension of Q.
Now any element g of G has finite order. Hence the eigenvalues of g are roots of unity. Moreover, the characteristic polynomial of g has coefficients in the field K; since its roots are algebraic numbers the coefficients are in L. Thus the eigenvalues are roots of unity satisfying a polynomial of degree n over L; hence if d is the degree of the field extension L of Q we have roots of unity of degree at most n . d over Q. There are only finitely many such roots of unity. Thus the order of G is bounded.