MrAl's Brief explanation:
notes:
Q1(c) means "Q1 collector"
Q1(ec) means "Q1 emitter collector"
Q2(ce) means "Q2 collecter emitter"
etc.

The switch is closed, current flows through Q1 (eb) turning it on somewhat.
Q1 turning on turns on Q2 somewhat, which causes its collector voltage
to fall somewhat, which pulls that end of C1 down which causes more current
to flow through Q1(eb) which causes more current to flow through Q2(eb) which pulls
Q2(c) down to about 0.2 volts (vsat). Also, C1 charges up with (-) at Q2(c).
Also, L starts charging which causes current to build through Q2(ce).

C1 charges quickly, which leaves the only drive to Q1 from R1. Since R1
doesnt drive Q1 all that much, when the collector current of Q2 rises to a
high enough level the base current supplied from Q1 times the Q2 beta
reaches the value of collector current and Q2 starts to come out of
sat. Once Q2 starts to come out of sat, its collector voltage starts to rise
so the voltage at the end of the cap at Q2(c) also starts to rise simply
because they are connected together. As it rises, the other end of the cap
also rises (C acting like a battery for a very short period of time) cutting off
Q1 because that end of the cap rises above the Q1 emitter voltage. Once
Q1 cuts off, Q2 also cuts off, and the Q2 collector rises to the voltage
determined by the nonlinear resistance of the LED times the current
now available from L which comes out to around 4.2 volts or so. The LED
thus gets a higher voltage then the input supply voltage. Now L dumps
its energy into the LED untill the cap discharges (through R1) which causes
the Q1 base voltage to become less then the Q1 emitter voltage and so
Q1 starts to turn on again.

While L dumps its current through the LED, the voltage
across the LED falls only slightly because of the nonlinearity of the LED,
but because Q1 is switching on and off the voltage across and the current through the
LED are only present for about 60% of the time. Averaging the current, it comes out
to about 40ma. In other words, if you were to supply about 40ma continuous uninterrupted
current flow through the LED you would get the same level of light output from the LED
as you do when pulsing it at a higher current which averages 40ma.
The reason for the pulses are so that the circuit can reset the inductor.
The inductor "kicks back" to provide the increased voltage and current which
could not otherwise be acheived with just two AA batteries.

Reviewing the above it can be seen that the charge time of the inductor
is determined mostly by the value of L and the transistor betas and the 3500 ohm resistor
and the battery voltage,
and the dump time of the inductor is mostly determined by the R1*C time constant,
although there is some interaction between these two.

----------------------------------------------

Numerical analysis:

A preliminary approximate numerical analysis was done on the circuit shown using
the values:
C=680pf, L=100uh, vsat=0.00v

Numerical results:
1. Transistor on time calculated: 4.58us (measured 4us)
2. Transistor off time calculated: 5.69us (measured 6us)
3. Max inductor current set to: 140 ma (set as measured)
4. Min inductor current calculated: 32 ma (measured 30ma)
5. Total oscillation period: 10.27us (measured 10us)

The differences are probably due to the estimation of the v/i curve of
the LED, which was approximated as:
v2=4.2*(1-exp(-80*i))+i {for i=0 to 140ma} {in the program i is the inductor current X2 and v2 is the LED voltage}
when the transistor Q2 is off and
v2=0.0 {Vsat of Q2}
when the transistor Q2 is on (v2 is the voltage across the LED).
The curve of the LED is critical to the analysis, as this basically
helps determine the minimum current through the inductor when the
transistor is off, so a better approximation can be used which would
improve the analysis in general. For now, we use the above and limit the current to
a max of 140ma (as measured with a battery voltage of 2.35 volts when the flashlight is on).
Note almost all the inductor current flows through the LED when the transistor is
off, hence the LED current is made equal to the inductor current when the transistor
is off. A very insignificant amount flows through the capacitor during this off time.
In the actual circuit the LED current peaks at 300ma for a very short time.

The waveform shown along with the schematic is the current through the LED as viewed
on an oscilloscope.

One last note: if the LED were powered by the two AA batteries in series
alone, the LED might light up, but very very dimly.