Why is torque assumed to be constant? Consider a spanner:

I was doing calculations on how much torque would be applied on the bolt. To do this, I considered how much force was being applied at the full length of the handle. Then, I equated this torque generated with the torque applied at the bolt.
$$\tau = Fd$$
$$\tau_{handle} = \tau_{bolt}$$
Very basic stuff that I've done over and over again. But it lead me to wonder... why exactly are these two torques assumed to be equal (line 2)?
I've never seen an explanation or proof for it. Can someone provide an explanation for why this is true?
I realised that whatever the reason is, it must mean for some reason, along this rigid body, the force becomes larger and larger, as the distance $d$ gets smaller and smaller (to maintain constant torque). I can't really see how our force is able to be amplified by this system.
Can someone explain to me why the torque can be transmitted without decreasing over distance? I'd prefer an answer aimed more at the intuition.
 A: To get to your second equation, we can consider the torque on just the spanner to start with.
The sum of the clockwise and anticlockwise torques on the spanner must be zero (otherwise it would start rotating).
So the $Fd$ torque is balanced by another torque from the nut, (acting on the spanner), it's caused by contact forces between nut and spanner $F'$ and the torque is $2F'r$, where $r$ is the radius of the nut.
These torques are equal.
So if the spanner doesn't start to rotate
$$Fd = 2F'r$$
Then considering the sum of torques on the nut - there are equal and opposite forces acting on the nut.  If the nut doesn't start to rotate the $2F'r$ torque is balanced by another torque caused by the friction force between the nut and bolt $F''$, perhaps this is what you call $\tau_{bolt}$.  This torque is from a friction force acting at an even smaller radius, the radius of the bolt, $r'$.
$$2F'r = F''r'$$
So putting the equations above together you get to your $$\tau_{handle} = \tau_{bolt}$$
$$Fd  = F''r'$$
From above the friction force between the nut and bolt must increase, for small bolts (if it is to resist the turning moment).  It depends on the radius of the bolt $$F'' = \frac{Fd}{r'}$$
So if $F$ or $d$ are large enough, or $r'$ is small enough to make the right hand side bigger than the maximum friction force between the nut and bolt, then the nut can be undone.
