There's a lot of traffic in this thread about powering the climber. Why
can't it simply have a diesel/gasoline engine with its own oxygen
supply?

just possibly something to this...

Or run electrical cables up the elevator to power an electric motor?

But this was a bad idea.

Why make things more complicated than they need to be?

Your proposal is not _totally_ implausible. The energy required to climb
a beanstalk is only a small fraction of the energy required to accelerate
a payload into Low Earth Orbit; the fuel and oxygen tankage required
would be large, but not prohibitively so.

I did a quick calculation assume modern diesels assuming a 10 tonne payload
starting with 10 tonnes of fuel. From my matlab script:

G=6.67300e-11; %gravitional constant
Mc=10e3; %Mass of cargo, 10tonnes
Me=5.95e24; %mass of earth
e=0.5; %efficiency of typical 4 stroke
Esp=40e6; %fuel energy per kg (assume air breathing)
r0=12760e3/2; %start point (radius of earth)
Mf=10e3; %mass of fuel

Equating the energy required to get out of the gravity well to the energy
available in the fuel tank:
GMe(Mf+Mc)(1/r0-1/r1) = e Esp Mf

This makes three assumptions, One is that we're air-breathing all the way,
which might not be too unreasonable because the hardest part (where g is
highest and the engine will be working its hardest) has plenty of air. Even
at high altititudes a super-charger or similar technology could be applied
to allow the engine to breath the air, only switching to a liquid oxygen
tank, or oxygen-less fuel when truely in space.
The second is that the mass of the fuel in the tank remains constant -
obviously it would be used up.
The third is that the 10tonnes of payload would of course include the heavy
engine.

I get from this about 1000km of altitude - only about 1/30th of the altitude
needed. So perhaps not entirely unreasonable, but difficult nonetheless.


Rest of matlab script:

r1=1/(1/r0 - (Esp*e*Mf)/(G*Me*(Mf+Mc))); %calculate height

altitude=(r1-r0)/1e6 %million meters (need about 30e6)