Quick questions on Mathematical induction for divisibility: standard technique and algebraic restructuring

To prove $E(n)$ is divisible by $d$ for all positive integers $n$, follow the same four parts as every other induction:

The add-and-subtract trick is the reliable way to manufacture the $E(k)$ chunk inside $E(k + 1)$. You start from $E(k + 1)$, and you add and subtract whatever is needed to make $E(k)$ appear, so that the expression splits into "$E(k)$" plus "a leftover", and you then check the leftover is a multiple of $d$. Because you add and subtract the same quantity, you have changed nothing, only the grouping.

For an expression with a single exponential like $5^n$, the fastest route is the multiplier method: pull out one factor of the base to step the exponent up, then substitute the hypothesis. From $E(k) = 5^k - 1 = 4M$, rearrange to $5^k = 4M + 1$, then

If the statement says "for all $n \ge 2$" or "for all $n \ge 0$", start the base case at the smallest valid $n$ and run the step from there. The inductive step still goes from $k$ to $k + 1$ exactly as before; only the base value changes.

$(k + 1)^3 = k^3 + 3k^2 + 3k + 1$, not $k^3 + 3k + 1$. A wrong coefficient destroys the factor of $d$.