In order to understand radioactive decay, we have to see what is in the nucleus of an atom and how its parts interact.
As you may know, all nuclei, no matter what kind of atom (element), consist of protons and neutrons (nucleons). A proton is a positively charged heavy body, and a neutron is just slightly heavier than a proton and has no charge. The neutron can be thought of as a proton with an electron (or, more accurately, beta particle) bound to it by the weak nuclear force.
The protons of a nucleus repel one another very strongly through the electric force of repulsion between like-charged bodies. However, the attractive strong nuclear force is much stronger than the electric force at such small distances, and so the strong nuclear force overcomes the electrostatic (Coulomb) repulsion and holds the protons and neutrons together.
This union results in a ball of protons and neutrons (to a certain approximation) shaking back and forth violently within the nucleus, but held together with the strong nuclear force. Sometimes, the nucleus' (or, isotope's) configuration (shape) is always energetically "stable," and will never break apart no matter how much time passes - like an intact water balloon (so long as something strong enough doesn't come along and break it). Other nuclei (of a different isotope), every so often in this violent vibration, assume a shape that cannot be held by the volumetric and surface tension. These are called "unstable" nuclei, or "radioactive" nuclei. When this happens, a piece of the nucleus, a particle, breaks off. This is called "decay," or "radioactive decay." When the nucleus decays, it does not disappear. It simply breaks into multiple pieces.
We cannot say, for any given "unstable" nucleus, exactly when it will assume a configuration leading to decay. But, if we have a large number of nuclei (say, N = 10^23; around the number of atoms when amassed you can see with your naked eye) we can say approximately that the number of decaying nuclei at time t, dN(t), must be proportional to the number of nuclei present at time t, N(t). Additionally, we can say that dN(t) should be proportional to the time that passes over a small enough duration, dt. Note that in these proportionalities N is considered continuous rather than discrete. This is an approximation - or, really, an error - since we cannot really have a fraction of a radioactive particle, by definition.
Continuing, nonetheless, we have said dN ~ N * dt.
Rearranging, and introducing a proportionality constant, L (the "decay constant"), we see
dN(t) / dt = - L * N(t)
The constant L is considered positive by definition, so the negative is introduced to capture that the change in the number of nuclei is negative over time period dt.
Rearranging the equation gives
dN(t)/N(t) = -L * dt
The solution to this equation through basic calculus is
N(t) = N(t=0) * exp(-L * t)
This is where the "e" comes from.
Now, your question is "does decay happen every second?" The problem is, the question assumes there is a yes or no answer. Also, it is useful to explain the half-life.
The half-life, by definition, is the time during which half of the particles in the sample should have decayed (become something else... not disappeared).
We can calculate this time by using the equation above as
N(t)/N(t=0) = 1/2 = exp(-L * h)
where h = half-life. Solving the right side, we get
-ln(2) = -L * h
or
h = ln(2)/L
This shows the relationship between the half-life and the decay constant. Over the time h, half of the original nuclei will have decayed, leaving N(t=0)/2 nuclei in their original state. After another half-life, half of the remaining undecayed nuclei will have decayed, with N(t=0)/4 remaining originals. Generally, after H half-lives, N(t=0)/2^H nuclei will remain undecayed.
The problem with the above derivation, as mentioned, is that it calculates an average behavior of a large number of nuclei (or, more accurately, a proportion of total nuclei) that, as far as the equation is concerned, is continuous. This is called the "classical" approach.
In order to derive the actual decay of a number of nuclei N, we should start with the statistical representation, which correctly treats the number of nuclei as discrete, but - since we only know the probability that any given nucleus will decay in a certain duration - gives a probability distribution of final outcomes instead of a deterministic result. Therefore, the answer to your question is: "In each second, there is a probability P that decay will happen, and there is a probability (1-P) that decay will not happen." Of course, once all nuclei have decayed, the probability P is zero.
It is possible, albeit exceedingly improbable for large numbers of nuclei, that in a sample of radioactive nuclei, all of them will decay in the same short time dt. We can call this outcome #1. There is only one way that this outcome can be achieved. If the probability that a nucleus decays in time dt is p, then the probability it does not decay in the same time is (1 - p). The probability of outcome #1 is p^N. There are N ways in which outcome #2 occurs, where only one particle decays in time dt. This means the probability of outcome #2 is (N!/(N-1)!)(1-p)p^(N-1) = N(1-p)p^(N-1). Outcome #3 is that 2 nuclei decay in time dt. The probability of outcome #3 is (N!/(N-2)!2!)(1-p)^2p^(N-2) In general, Outcome #k, that k-1 nuclei decay in time dt has probability (N!/((N-(k-1))!(k-1)!))*(1-p)^(k-1)*p^(N-(k-1)).
One of all of these outcomes must be fulfilled at the end of time dt, so the summation of all these probabilities is one.
I will leave the remainder to your derivation.