how does difficulty work exactly?

[Shogen]

If BTC falls into this situation are there any plans to keep blocks moving at 1 per 10 mins if a crazy drop in mining power would happen and average block time goes into weeks?

Currently, no.
I seriously doubt if majority of hashrate will left bitcoin all in a sudden, unless in an huge disaster causing world-wide internet and electricity outage.
I also doubt if the difficulty algorithm will be adjusted, as it will lead to hard fork.

[durrrr]

i dont think that the difficulty will ever be lower than it is at any point in time. meaning it only gets harder because it is growing and more poeple are adopting and to spread the wealth more the more users needed mining and supporting the network and the network gets bigger

[JoelKatz]

i dont think that the difficulty will ever be lower than it is at any point in time. meaning it only gets harder because it is growing and more poeple are adopting and to spread the wealth more the more users needed mining and supporting the network and the network gets bigger

There are a lot of marginal miners now with older miners that make just a bit more than the cost of the electricity needed to run them. If the price of Bitcoin drops significantly, they'll stop mining. However, even though there are a lot of them in number, they probably don't represent a huge fraction of the mining power out there. Still, it wouldn't surprise me if difficulty temporarily dropped a bit if the price of Bitcoin, say, dropped by 30%.

[counter]

huh, I always just imagined the code would be come more complex and harder to solve therefore making the difficulty higher then before.  This is a much more complex question then I'd assumed.

[Nagato4]

huh, I always just imagined the code would be come more complex and harder to solve therefore making the difficulty higher then before.  This is a much more complex question then I'd assumed.

Indeed the code is pretty simple.
It just check the time taken for finding the 2016 blocks and update the "target" value accordingly.