we know that φ^{n}is onto.

for suppose a is any element of A. since φ is onto, there exists b_{1}with φ(b_{1}) = a.

since b_{1}is likewise an element of A, there exists b_{2}with φ(b_{2}) = b_{1}, therefore φ^{2}(b_{2}) = φ(φ(b_{2})) = φ(b_{1}) = a.

a short inductive proof then shows that φ^{n}is onto, for any n.

so if a is any element of the kernel of φ (so that φ(a) = 0), we know there exists b in A with φ^{n}(b) = a, no matter what n we pick.

in particular, if we pick n = N (the integer at which the kernels stabilize), then a = φ^{N}(b).

in which case φ^{N+1}(b) = φ(φ^{N}(b)) = φ(a) = 0.

so if a is in ker(φ), then b is in ker(φ^{N+1}) = ker(φ^{N}). which means φ^{N}(b) = 0, that is: a = 0, so φ is injective.