Proof by Induction: Puzzle Pieces Problem There's a thought puzzle I am struggling to understand that deals with the fundamentals of writing a proof involving the inductive assumption.

A jigsaw puzzle is solved by putting its pieces together in the correct way. Show that exactly $n−1$ moves are required to solve a jigsaw puzzle with $n$ pieces, where a move consists of putting together two blocks of pieces, with a block consisting of one or more assembled pieces.

So, a proof by PMI would look something like this:
I will verify this hypothesis is true for at least one value of $n$:
Consider:
pieces: $n=1$ and fits: $n-1=0$ (valid)
pieces: $n=2$ and fits: $n-1=1$ (valid)
pieces: $n=3$ and fits: $n-1=2$ (valid)
I will assume the hypothesis holds true from $n=3$ up to some arbitrary value $k$ ie. $k$ pieces will have $k-1$ fits. I will now prove true for $k+1$ pieces to show $k$ fits.
It is at this part of the proof that confuses me. Clearly, during the last fit, the two subsections of the puzzle, arbitrary denoted as $p$ and $q$ must comprise the $k+1$ pieces as part of the inductive assumption. Also, each subsection also took $p-1$ fits and $q-1$ fits respectively, but I don't really know how to combine all this knowledge together to mathematically show $k$ fits.
Any help would be greatly appreciated! Thanks!
 A: By definition, notice that $p + q = k + 1$. By the induction hypothesis, the last two blocks required $p - 1$ and $q - 1$ fits, respectively. Adding in the last fit, we conclude that the total number of fits is:
$$
(p - 1) + (q - 1) + 1 = (p + q) - 1 = (k + 1) - 1 = k
$$
as desired.
A: Here's another way to construct a proof using strong induction.
Let $P(n)$ be the statement: A jigsaw puzzle is solved by putting its pieces together in the correct way. Show that exactly $n−1$ moves are required to solve a jigsaw puzzle with $n$ pieces, where a move consists of putting together two blocks of pieces, with a block consisting of one or more assembled pieces.
Base Case. For $P(1)$ we see that no moves are required for a jigsaw puzzle with $1$ piece since it is already constructed and thus $1-1=0$ moves are required and our proposition holds.
Inductive Step. Assume that $1\le j \le k$ and $k\ge 1$ and that $P(j)$ is true. 
Let's say we have a puzzle with $k+1$ pieces. Then it must be the case that the final move to construct our puzzle consisted of putting together a block consisting of $j$ pieces and a block consisting of $k+1-j$ pieces. We see that each of $j$ and $k+1-j$ are between $1$ and $k$ inclusive so by the inductive hypothesis the blocks of $j$ and $k+1-j$ pieces took $j-1$ and $k+1-j-1$ moves to put together, respectively. So the total number of moves to put together our $k+1$ piece puzzle was our one final move plus all prior moves which means that it took $1+(j-1)+(k+1-j-1)=k$ moves. But then a puzzle of $k+1$ pieces took $(k+1)-1=k$ moves to complete and we have shown that $P(k)\Rightarrow P(k+1)$.
It follows by strong induction that $P(n)$ is true for all non-negative $n$. $\Box$
