Prove: If each of $m_1,\dotsc,m_n$ can be written as sums of two squares, so can their product $m_1 \dotsm m_n$ The Problem

Prove: If each of $m_1,\dotsc,m_n$ can be written as sums of two
  squares, so can their product $m_1 \dotsm m_n.$ (Hint: Use induction.)

(I believe that $m_1\dots m_n$ represent composite numbers in this case...)

Some Background
This problem is actually the third part of a multistep problem.  The previous step had us do the following:

Prove: If $m_1$ and $m_2$ can be written as sums of two squares, so
  can their product $m_1 m_2$.

To which I offered the following proof:
Proof.
Suppose $m_1$ and $m_2$ can be written as sums of two squares, where $m_1=a^2+b^2$ and $m_2=c^2+d^2$.  Then the product $m_1m_2=(a^2+b^2)(c^2+d^2)$.  Having just showed that $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$, we have $m_1m_2=(ac-bd)^2+(ad+bc)^2$.  Letting $j=ac-bd$ and $k=ad+bc$, we have $m_1m_2=j^2+k^2$, which is the sum of two squares. $\square$ 
(The part where I used $(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$ was proven in the very first step of this problem.)

My Questions
I feel that this question shouldn't be much more difficult than the previous step, but I am failing to see how to apply induction here, and thus have been struggling to see how I should even start the problem off.  More specifically,


*

*Does the previous step count as the base case for this step, with $n=2$?  If not, what should my base case look like?

*What exactly should my inductive step look like here?



As always, your help is very appreciated!
 A: *

*Sort of, the previous step acts as both a proof for the base case (with $n=2$) and as the basis for the inductive step. 

*You assume that "if each $m_1, \ldots, m_n$ can be written as a sum of two squares then so can their product". You want to show that this assumption implies "if each $m_1,\ldots,m_{n}, m_{n+1}$ can be written as a sum of two squares, then so can their product.". 
This isn't as hard as it looks. You know that $m_1m_2\cdots m_n$can be written as a sum of squares by using your induction hypothesis. You also know that $m_{n+1}$ can be written as a sum of two squares. 
So can $$\underbrace{m_1 m_2 \cdots m_n}_{\text{sum of 2 squares}} \cdot \underbrace{m_{n+1}}_{\text{sum of 2 squares}}$$ be written as a sum of two squares? (think back to the previous part now, consider $m_1\cdots m_n$ and $m_{n+1}$ as your two numbers). 
A: You are right. "The previous step" covers the induction case, since we are dealing with two numbers $m_1,m_2$ over there.
For induction, consider $k+1$ numbers $m_1, ...,m_{k+1}$ that can be written as the sum of two squares.
In the inductive step, you can assume that the product of $m_1, m_2, \ldots,m_k$ can be written as a sum of two squares (since here we can assume that for all integers $< k+1$,  the premise is true).
Now, all you need to do, is to apply step two (or the inductive case $k=2$), with the two numbers $m_1 ... m_k$ (the product) and the last number $m_{k+1}$. That tells you that their product, which is $m_1 ... m_{k+1}$, is writable as a sum of two squares, which was to be proved.
So you were right, nothing more than step $2$ is required for the proof of the third part.
