Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures. 
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or $n+2$ digits of the form

$$\underbrace{4\cdot a}_{1\text{ or }2}~ \underbrace{aaa\ldots a}_{n-2} ~\underbrace{7\cdot a}_{2^{*}}.$$

$*$ if $a=1$, then $7 \cdot a$ is the sequence of digits $07$. 
I am wondering whether the result above can be proven using some number-theory tool.
Thanks in advance!

Observe the patterns in the written calculation of these products
$$a=1\to\begin{matrix}3&7\\&3&7\\&&3&7\\&&&3&7\end{matrix}$$
$$a=2\to\begin{matrix}7&4\\&7&4\\&&7&4\\&&&7&4\end{matrix}$$
$$a=3\to\begin{matrix}1&1&1\\&1&1&1\\&&1&1&1\\&&&1&1&1\end{matrix}$$
$$a=4\to\begin{matrix}1&4&8\\&1&4&8\\&&1&4&8\\&&&1&4&8\end{matrix}$$
$$a=5\to\begin{matrix}1&8&5\\&1&8&5\\&&1&8&5\\&&&1&8&5\end{matrix}$$
$$\cdots$$
$$a=9\to\begin{matrix}3&3&3\\&3&3&3\\&&3&3&3\\&&&3&3&3\end{matrix}$$
A digit of the product, in a complete column, is the sum of the digits of $37\cdot a$ plus a possible carry from the previous column, i.e. the sum of the digits of the sum of the digits of $37\cdot a$.
The initial and final digit pairs correspond to partial sums.
We have
$$
37\cdot1=37\to10\to1,\\
37\cdot2=74\to11\to2,\\
37\cdot3=111\to3\to3,\\
37\cdot4=148\to13\to4,\\
37\cdot5=185\to14\to5,\\
\cdots\\
37\cdot9=333\to9\to9.$$
The phenomenon is explained by the fact that $37=4\cdot9+1\to10\to1$, and this occurs for all factors of the form $k\cdot9+1$. For example,
$$73\cdot66666666666666=4866666666666618.$$
