Poisson and exponential distributions

hello,

i'm having some trouble seeing the relationship between the Poisson and the
exp. distributions. specifically,

if the Poisson with rate L and time t is:

P(k| L,t) = (Lt)^k exp(-Lt) / k!

then i think that the exp. process with rate-parameter L, should basically
be P(0| L, t) = exp(-Lt). why doesn't the normalizing constant (1/L) come
out directly from this to give Lexp(-Lt) ?
does this have something to do with the fact that P(0| L, t) as above is a
likelihood and not a probability ? if so, then using laws of probability,
can the exponential distribution be derived, i.e.:

P(t|0,L) = p(0|L,t)p(t|L) / p(0|L)

conversely, how would one derive the poisson from the exponential
distribution ?

thanks,
-g


.

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