Congratulations on deriving the exponential law for yourself, one learns a great deal about science working like this. Now to your last question:
If I had a group of atoms that have an 'average lifetime' of say 5 seconds, after 5 seconds has elapsed, what is the 'average lifetime' of the remaining atoms? I don't think I can arbitrarily choose some reference time to begin ticking away at the atoms' remaining time, does that mean at any point of time that their 'average lifetime' or expected lifetime is always a constant, and never actually diminishes as time goes on?
Yes indeed the average lifetime is constant. And the exponential distribution you have derived is the unique lifetime distribution with this property. Another way of saying this is that the decaying particle is memoryless: it does not encode its "age": there is nothing inside the particle that says "I've live a long time, now its time to die". Yet another take on this - as a discrete rather than continuous probability distribution - is the geometric distribution of the number of throws before a coin turns up heads, and the observation that a coin has no memory that counters the famous gambler's fallacy.
To understand this uniqueness, we encode the memorylessness condition into the basic probability law
$$p(A\cap B) = p(A) \, p(B|A)$$
Suppose after time $\delta$ you observe that your particle has not decayed (event $A$). If $f(t)$ is the propability distribution of lifetimes, then the probability the particle has lasted at least this long, i.e. the probability that it does not decay in time interval $[0,\,\delta]$ is:
$$p(A) = 1-\int_0^\delta f(u)du$$
The a priori probability distribution function that the particle will last until time $t+\delta$ and then decay in the time interval $dt$ (event $B$) is
$$p(B\cap A) = f(t+\delta) dt$$.
This is events $B$ and $A$ observed together, which is the same as plain old $B$ since the particle cannot last unti time $t + \delta$ without living to $\delta$ first! Therefore, the conditional probability density function is
$$p(B|A) = \frac{f(t+\delta)\,dt}{1-\int_0^\delta f(u)du}$$
But this must be the same as the unconditional probability density that the particle lasts a further time $t$ measured from any time, by assumption of memorylessness. Thus we must have:
$$\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta),\;\forall \delta>0$$
Letting $\delta\rightarrow 0$, we get the differential equation $f^\prime(t) = - f(0) f(t)$, whose unique solution is $f(t) = \frac{1}{\tau}\exp\left(-\frac{t}{\tau}\right)$. You can readily check that this function fulfills the general functional equation $\left(1 - \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$ for any $\delta > 0$ as well.
As Akhmeteli's answer says, true memorylessness is actually incompatible with simple quantum models. For example, one can derive the exponential lifetime for an excited fluorophore from a simple model of a lone excited two state fluorophore equally coupled to all the modes of the electromagnetic field. The catch is that the derivation rests on approximating an integral over positive energy field modes by an integral over all energies, both positive and negative. This of course is unphysical, but an excellent approximation since only modes near to the two state atom's energy gap will be excited: the fluorophore "tries" to excite all modes equally, but destructive interference prevents significant coupling to modes of greatly different energy than the difference between the energies of the states on either side of the transition.
I show how this analysis is done in this answer here and here.
Linewidths are mostly extremely narrow compared to the frequencies of the photons concerned, so I find it surprising and quite wonderful that Ahkmeteli cites a paper giving experimental evidence of the nonconstant lifetime.
Let's say $\lambda = 0.5 s^{-1}$. So we would intuitively expect half of the atoms in a population to have decayed after 1 second.
Those don’t follow. To model a system where half the atoms decay after one second, use the half-life $\tau_{1/2} = 1\,\rm s$, and the relation
$$
N(t) = N_0 \cdot 2^{-t/\tau_{1/2}}
$$
This is the same as your relation if
\begin{align}
e^\lambda &= 2^{1/\tau_{1/2}}
\\
\lambda &= \frac{1}{\tau_{1/2}} \ln 2
\end{align}
For a half-life $\tau_{1/2}=1\rm\,s$, this relation gives $\lambda = 0.69\rm\,s^{-1}$. You are underestimating the decays because your decay constant is too small.
Best Answer
The first answer (form Shaktal)is in my opinion perfect from the mathematical point of view (solving the differential equation) as well as from the physical point of view (having all units in order).
However your probably conceive your approach more intuitive because you know this from calculating things with Interest i.e.
$N_{Money}=N_{0,Money}\times (1\pm Interest)^{t(in\;years)}$
Aside from the units being messed up, physicists like functions like $e^{something}$ because you can easily calculate with them. I.e. try to take $\frac{d}{dt}N_{wrong}$.
$\frac{d}{dt}N_{wrong}=\frac{d}{dt}N_0\times(1-k)^t=\frac{d}{dt}N_0 e^{ln(1-k)t}$
gives you just
$\frac{d}{dt}N_0 e^{-\lambda t}$
so why the detour?
The derivate is called the activity and is very important so you have to calculate it quite often. You may also want to integrate $N_{wrong}$ sometimes which would also be really annoying.
Further considering the answer from Samuel Hapak: The half-life (not the great game) as the "real" half-life of an element, as in $N_{1/2}=N_0\times\left(\frac{1}{2}\right)^{t_{1/2}/\lambda}\;\;t_{1/2}\text{: half-life, }\lambda:\frac{1}{\text{(half-life)}} $
gives you a quick overview over the decay rate with an SI-unit. So it can easily be given an order of Magnitude one can relate to. So it is indeed as intuitive as it can get in my opinion. One can of course easily switch to the mathematical expression resulting from solving the differential equation shown in Shaktal's answer:
$N_t=N_0 e^{-\tilde\lambda t}$
by setting $\tilde\lambda = ln(2)\lambda $