To answer this question it is helpful to draw a diagram of the forces acting on the aircraft whilst it is in the climb, and resolve them accordingly.

If you do this, you can see that the lift vector is acting at 90 degrees to the flight path. (The flight path being at an angle to the horizon)

However, the weight vector will always act straight down, or at 90 degrees to the horizon. So only a part of this weight vector is acting in the direction opposite to the lift. (The other part of the weight vector acts in the same direction as the drag)

Therefore, in order to keep the forces balanced, so the aircraft isn't accelerating:

Lift = The part of the weight vector acting perpendicular to the flight path.

To be completely mathematical

L=W cos(gamma); where gamma is the flight path angle to the horizontal.

So you can see that lift is now a fraction of the weight, so lift is less than weight. And mathematically this is shown because cos(gamma) can only have a value between 0 and 1. With it equalling 1 when gamma is zero (ie horizontal flight) and 0 when gamma is 90 degrees (ie full on vertical climb)

The load factor bit comes from the definition of load factor, which is Lift/Weight. Now we've already determined that Lift is less than Weight in steady climb, so this fraction has a numerical value less than one.

Again being mathematical, you can say from Lift = Weight x Cos(gamma)

Lift/Weight = Cos(gamma)

hence

Load factor = Lift/Weight = Cos (gamma)

And once again you can see that load factor can range from zero to one, with it equalling one, only in straight and level flight.