Let's assume the magnetic field vectors point in z-direction (or: let's call the direction the magnetic field vector points "z"). Then we have for the magnetic field:
$$\vec{B} = \begin{pmatrix}0\0\B\end{pmatrix}$$
and for the speed of the electron:
$$\vec{v} = \begin{pmatrix}v_x\v_y\v_z\end{pmatrix}$$
The Lorentz-force $\vec{F}$ due to a magnetic field is given by
$$\vec{F} = q\vec{v}\times\vec{B}$$ $$= q\begin{pmatrix}v_x\v_y\v_z\end{pmatrix}\times\begin{pmatrix}0\0\B\end{pmatrix}$$ $$= q\begin{pmatrix}v_y B - 0\0 - v_x B\0 - 0\end{pmatrix}$$ $$= qB\begin{pmatrix}v_y \ - v_x \0 \end{pmatrix}$$
So as you can see the force in z-direction is 0 and due to <PERSON> law ($F = m a$) the acceleration is as well, resulting in constant speed in z-direction.
If you now solve the differential equations
$$\frac{d}{dt}\begin{pmatrix}v_x \ v_y \v_z \end{pmatrix} = qB\begin{pmatrix}v_y \ - v_x \0 \end{pmatrix}$$
you will find that the path is indeed helical, i.e. that $$\vec{v}(t) = \begin{pmatrix}v_0 \sin(\omega t + \phi) \ v_0 \cos(\omega t +\phi) \v_z \end{pmatrix} $$ is a valid solution. 418
(Disclaimer: might not be the only possible solution) $\omega$ here depends on q and B, while $\phi$ should depend on your coordinate-system and $\vec{v}(0)$.
Edit:
For the differential equation you could do this:
$$\frac{d}{dt} v_x = qBv_y$$ $$\frac{d^2}{dt^2} v_x = qB\frac{d}{dt}v_y = -q^2B^2v_x$$ $$\frac{d^2}{dt^2} v_x = -q^2B^2v_x$$
So we are searching for a function whichs second derivative is the function itself but with a negative factor. So you either look it up in a table, do some fancy math-magic, assume it's an exponential function or just know from previous studies that its a linear combination of $\sin$ and $\cos$. Linear combinations of sin and cos can be represented as a phase $\phi$, so pick one randomly, add a phase, calculate $v_y = \frac{d}{dt} \frac{v_x}{qB}$ and you are done.
For a mathematical proof that this is the only solution or additional solutions read the pack insert and ask your doctor or mathematician. ;-) 418