Electronic – Default-on, latch-off, MOSFET power switch

I'd like to build a low-battery circuit. It should default on when first connected, but if a particular signal goes high for a small amount of time (say, 10 microseconds,) then it should turn off, and stay off.
It's the "default on" that gives me trouble. The closest I can get is something like this:

^{simulate this circuit – Schematic created using CircuitLab}

From null, "OFF" is floating. Supply 15V, and the capacitor starts out acting like a bit of wire, which pulls the M1 N-channel gate high, which makes the N-channel conduct (I'm likely to use a BS170, not the IRF on the schematic.) That pulls the gate of the P-channel M2 low, which makes it conduct. That, in turn provides voltage into R3/R2 divider that keeps the circuit latched open.

Now, when OFF goes low-impedance to ground, it will turn off M1, and thus R1 will pull the M2 gate high, and M2 will turn off. Because C1 already has a charge, the system will now stay with M2 turned off even if the OFF signal goes high impedance again (which it will, because it's a MCU output eventually powered by the SWITCHED voltage.)

At this point, the system is unlikely to turn on again unless I short "OFF" to "+15V." In fact, I may add a pushbutton to do just that for manual turn-on.

While off, the leakage through C1 and the leakage through R1/M1 (and, I guess, leakage through M2 and the entire powered subsystem) will still draw a little bit on the battery, so if it's a LiPo, I shouldn't leave it in that state forever or it will die, but as a safeguard for a hobby robot, I think it might work.

Is this correct, or am I getting something wrong? Will the system oscillate when the "off" is triggered? Will the system automatically turn on from "zero" as intended?