Prove: $M(n+1) =(n+1)(1+M(n))$ with $M(n)$ being the number of multiplications required to find the determinant So I'm working on this problem where I have to show that to find the determinant of an $n \times n$ matrix you would require more multiplications than $n!$ as long as $n \geq 2$ and multiplications with $\pm1$ do not count. 
There are 3 parts I need to prove, and I've done 2 of them, the first and 3rd but I'm stuck on the 2rd. 
The function $M(n)$ is defined as the number of multiplications required to find the determinant of an $n \times n$ matrix, using Laplace expansion and excluding the multiplication with the $\pm1$ (the alternating sign before each expansion).
First part was to show that for a $1 \times 1$ matrix, $M(1) = 0$, meaning I require $0$ multiplications. I proved that using the definition of a $1 \times 1$ determinant. 
Second part, is where I am having a problem. I had to show that the recursive equation $M(n+1)=(n+1)(1+M(n))$ is valid for all $n \geq 2$, $n \in \mathbb{N}$. I tried proving this using induction, but I'm getting stuck because I always got to a point where I still had some version of the $M(n)$ function there which I couldn't get rid of. My thought process was that if I keep applying the recursion, eventually I'll get to $M(1)$ which I proved above is equal to $0$, and then through the "ripple effect" of the principle of recursion I can prove that the whole equation is valid, but I'm not exactly sure how to mathematically show that.
Third part was to show that $M(n) \geq n!$ for all $n \in \mathbb{N} \geq 2$. I proved this using the second part, although I still haven't proved the second part. I used induction and subbed the above equation for $M(n)$.
Any help on how to approach the second part would be appreciated. 
Best regards!
 A: If $A$ is square matrix with $n+1$ rows, $a_{ij}$ is scalar in $i$-th row and $j$-th column and $A_{ij}$ is corresponding minor of $A$, then Laplace expansion gives $$detA=\sum_{i=1}^{n+1}(-1)^{i+j}*a_{ij}*A_{ij}$$
It is expansion in respect to $j$-th column. $A_{ij}$ is determinant of an 
$n\times n$ sized matrix, so it takes $M(n)$ multiplications to compute every 
$A_{ij}$ by inductive hypothesis, so $a_{ij}*detA_{ij}$ needs one more, and there are $n+1$ summands that has to be computed separately. By neglecting multiplication of $(-1)^{i+j}$ this gives $M(n+1)=(n+1)(1+M(n))$.
A: But in part two, you do NOT need to get rid of $M(n)$ or get all the way down to $M(1)$. In this part, you only need to prove the recursive equation, not deduce an explicit formula for $M(n)$. As soon as you get an expression for $M(n+1)$ in terms of $M(n)$, you're done (with this part). And you don't need induction here either. Induction will be employed to deduce part three using the statement of part two.
I suspect that you may have already obtained a correct solution, but didn't recognize that you could stop. We can compare our notes. Expanding an $(n+1)\times(n+1)$ determinant along any row or column, for example along the first row, will give you $(n+1)$ terms of the form
$$\det(A)=\sum_{k=1}^{n+1}(-1)^{1+k}a_{1k}C_{1k}=a_{11}C_{11}-a_{12}C_{12}+\cdots\pm a_{1n}C_{1n}.$$
Each multiplication by $\pm1$ can be ignored. Each minor $C_{ij}$ of size $n\times n$ requires $M(n)$ multiplications, and there are $n+1$ of them. And then there are $n+1$ more multiplications of an element by an (already computed) minor. Adding together, this gives us the desired formula for $M(n+1)$ in terms of $M(n)$. That's it — for this part.
