Multiplication rule for independent sequential steps

Step 1 (base case $k=1$): with a single step there are trivially $N_1 = n_1$ ways, matching the formula for $k=1$.

Step 2 (base case $k=2$): for each of the $n_1$ ways to perform step 1, step 2 can still be done in $n_2$ ways, since its count does not depend on step 1's outcome. This splits all pairs of choices into $n_1$ disjoint groups of size $n_2$ (one group per outcome of step 1), so the addition rule gives a total of $n_2$ added to itself $n_1$ times, i.e. $N_2 = n_1 \cdot n_2$.

Step 3 (induction on $k$): assume the formula holds for $k-1$ steps, so the first $k-1$ steps together admit $N_{k-1}$ outcomes. Treat those $k-1$ steps as one combined "super-step" with $N_{k-1}$ outcomes, and step $k$ as a second, independent step with $n_k$ outcomes (its count still does not depend on earlier choices). Applying the two-step base case to this pair gives $N_k = N_{k-1} \cdot n_k$, which is $N_k = n_1 \cdot n_2 \cdots n_k$. By induction the formula holds for every $k$.