Step 1: Use dimensional and limiting behavior reasoning instead of memorizing the formula:
Consider what should happen physically as $N$ becomes very small, that is as almost all the nuclei have already decayed. With very few nuclei left, the number of decay events per unit time must also become very small, and in the limit $N \to 0$ the decay rate $-\frac{dN}{dt}$ must also approach 0, since there is almost nothing left to decay.
Step 2: Test option (A), $-\frac{dN}{dt} = \lambda N^2$, against this limit:
As $N \to 0$, $N^2 \to 0$ as well, so this form does go to zero, but it implies the decay rate falls off much faster than proportionally as $N$ decreases, which would require decay to depend on pairs of nuclei interacting. Radioactive decay of a single nucleus is a spontaneous, independent process, not a pairwise interaction, so this quadratic dependence does not match the known physical mechanism.
Step 3: Test options (C) and (D), which have $N$ in the denominator, against the same limit:
For option (C), $-\frac{dN}{dt} = \lambda / N^2$, and option (D), $-\frac{dN}{dt} = \lambda / N$, as $N \to 0$ the right hand side blows up to infinity instead of going to zero. This would mean the decay rate becomes infinitely large exactly when almost no nuclei are left to decay, which is physically impossible. Both options (C) and (D) are ruled out by this limiting argument.
Step 4: Test option (B), $-\frac{dN}{dt} = \lambda N$, against the same limit:
As $N \to 0$, $\lambda N \to 0$ smoothly and proportionally, matching the physical expectation that decay activity fades away in direct proportion to how much radioactive material remains. This linear form is also consistent with the well established exponential decay curve $N = N_0 e^{-\lambda t}$ observed experimentally for all radioactive isotopes.
Step 5: Conclude:
Only the first order linear relationship in option (B) is physically consistent, both from the underlying independent-decay mechanism and from the correct limiting behavior as $N \to 0$.
Final Answer:
\[ \boxed{-\dfrac{dN}{dt} = \lambda N} \]