The fundamental principle of counting states this: Suppose that two experiments are to be performed. Experiment $1$ can have $n_1$ possible outcomes and for each outcome of experiment $1$, experiment $2$ has $n_2$ possible outcomes, then together there are $n_1 \times n_2$ possible outcomes. Generalize to $k$ total experiments.
This seems like an excercise in strong induction so I will give it a shot, is it correct?
For $k=2$ the given statement is satisfied by the proposition above. Thus the base case is satisfied.
Assume that if for each $1 \leq i \leq k-2$, experiment $i$ can have $n_i$ possible outcomes and for each outcome of experiments $1,...,i$ experiment $i+1$ has $n_{i+1}$ possible outcomes, then there are $\prod\limits_{1\leq j \leq i+1}n_j$ possible outcomes of experiments $1,...,i+1$. Then for $k$ experiments if for each outcome of experiments $1 \leq i \leq k-1$, experiment $k$ has $n_k$ possible outcomes, then there are $(\prod\limits_{1 \leq i \leq k-1}n_i) \times n_k=\prod\limits_{1 \leq i \leq k}n_i$ possible outcomes altogether. So then by induction if $k$ experiments are performed;if experiment $1$ can have $n_1$ possible outcomes and for each outcome of experiments $1,...,k-1$ experiment $k$ has $n_k$ possible outcomes, then in all there are $\prod\limits_{1 \leq i \leq k}n_i$ possible outcomes.
