How to prove this modular multiplication property to be true? I am watching a youtube video on modular exponentiation https://www.youtube.com/watch?v=sL-YtCqDS90
Here is author's work

In this problem, the author was trying to calculate $5^{40}$
He worked the powers of 5 by 2 until he go to the largest exponent that was less than 40, 32. He then reasoned that $5^{40}$ mod 7 =  $5^{32}$ mod 7 * $5^ 8$ mod 7. Does anyone know where he got this property from?  I know that there is an exponent property that would state $5^{40}$ = $5^{32}$ * $5 ^ 8$ (from
http://hotmath.com/hotmath_help/topics/properties-of-exponents.html ) but can someone show how to get from that one to the mod version property? I tried looking it up but couldn't find it.
 A: The aspect you're puzzled by, $(5^{40} \bmod 7) \equiv (5^{32} \bmod 7)(5^{8} \bmod 7) \pmod 7$, is really the same process that has been used through the whole squaring process that preceded it, where (for example) we had  $(5^{32} \bmod 7) \equiv (5^{16} \bmod 7)(5^{16}\bmod 7) \pmod 7$.
Essentially it is the same property that you are probably familiar with from modular arithmetic in general. For example, consider calculating $(20\times 15) \bmod 7 $. We can take the $\bmod 7$ value of each: $20 \equiv 6 \bmod 7$ and $15 \equiv 1 \bmod 7$ - and use those values to calculate the result $(6 \times 1) = 6 \equiv 6 \bmod 7$. And to show that is truly the $\bmod 7$ value of the original multiplication, we can drop back into the $20=7k+6$ type  representation and prove out that all the multiples of $7$ can be ignored in the multiplication process. And if we want to in this case, we can even check that $300-6$ is indeed divisible by $7$.
So all we're looking at there is a standard property of modular arithmetic - the congruence classes multiply together consistently.

All numbers $n$ such that $n \equiv k \bmod m$ are in the same congruence class $\bmod m$ . For example, $\{2,9,16,23,30,37, \ldots\}$ are all in the same congruence class $\bmod 7$, identified by the non-negative member of the class less than than the modulo number - in that case, $2$. Still working modulo $7$, if we multiply any member of congruence class $2$ by any member of congruence class $4$, the answer will be in congruence class $1$.
A: That's the squaring step in the standard method of modular exponentiation by repeated squaring
$$\!\!\!\!\!\begin{align} a^{\large n}\ {\rm mod}\ m &\,=\, b\\ \Rightarrow\ \ \ a^{\large 2n}\ {\rm mod}\ m &\,=\, b^{\large 2}\ {\rm mod}\ m\end{align}\qquad $$
The proof is easy by converting to congruences - which are more convenient arithmetically 
$$\begin{array}{rrl} &a^{\large n}\ {\rm mod}\ m &\!\!=\ b\\ 
\Rightarrow&\ \     a^{\large n}\! &\!\!\equiv\ b\ \ \pmod m\\
\color{#c00}\Rightarrow&\ \     a^{\large 2n}\! &\!\!\equiv\ b^{\large 2} \pmod m\\
\Rightarrow&\ a^{\large 2n}\ {\rm mod}\ m &\!\!\!=\ b^{\large 2}\ \ {\rm mod}\ m\end{array}\qquad$$
where the $\rm\color{#c00}{red}$ inference employs the $ $ Congruence Power Rule for power $\,k = 2\,$ (squaring)
$$\qquad\qquad\ \  \begin{align} c &\,\equiv \, d\pmod m\\ \Rightarrow\ \ \ c^{\large k}\! &\,\equiv\, d^{\large k}\!\!\!\!\pmod m\end{align}\qquad $$
which is an inductive extension of the Congruence Product Rule $ $ (which could instead be used)
$$\qquad\qquad\ \  \begin{align} c &\,\equiv \, d\pmod m\\ \bar c &\,\equiv \, \bar d\!\pmod m\\ \Rightarrow\ \ \ c\bar c\! &\,\equiv\, d\bar d\!\!\!\!\pmod m\end{align}\qquad $$
The other inferences are standard conversions between mod as an operator and congruence, and their proofs are straightforward using the definitions.
Remark $\ $  Generally, as above, it is straightforward to prove properties of the mod operator by first converting to congruence form, then applying the arithmetical laws of congruences then, finally, converting back to operator form. While it is possible to perform such proofs without congruences, generally that will be messier, and less conceptual. 
A: Given:
$$A \bmod n = a => A = n * k_A + a$$
and
$$B \bmod n = b => B = n * k_B + b$$
then:
$$A * B = (n * k_A + a) * (n * k_B + b)$$
$$A * B = n^{2} * k_A * k_B + n * k_A * b + n * k_B * a + a*b $$
As $k_A$, $k_B$, $a$ and $b$ are integers, the modulo operation removes every multiple of n, leaving only $a * b$
therfore:
$$(A * B) \bmod n = (a * b) \bmod n$$
Given that exponation is a multiplication, this should explain the property you were asking for
A: The property you are asking about is why from $ab=c$ (in your concrete case with $a=5^8$, $b=5^{32}$ and $c=5^{40}$) one can deduce that $(a\bmod n)(b\bmod n)=c\bmod n$ for any positive integer $n$ (here $n=7$). Actually that is not right, since it boils down to $4\times 4=5^{40}\bmod 7$, which of course cannot be true, since $16$ is not a valid remainder modulo$~7$. What is going to be true is that the right hand side will be equal to $16\bmod 7=2$. Therefore the equation to establish should be formulated as
$$
  ((a\bmod n)(b\bmod n))\bmod n=(ab)\bmod n.
$$
Since that makes a lot of applications of "$\bmod n$", so it is more convenient to state it using the relation $x\equiv y\pmod n$, which one may define to hold if and only if $x\bmod n=y\bmod n$. Now obviously one has in particular $a\equiv a\bmod n\pmod n$, so it will suffice to prove the more general fact
$$
  a\equiv a'\pmod n \text{ and } b\equiv b'\pmod n \implies ab\equiv a'b'\pmod n
$$
(this is applied with $a'=a\bmod n$ and $b'=b\bmod n$). Now it is easy to see that $x\equiv y\pmod n$ if and only if $y=x+kn$ for some integer $k$. Therefore one may write $a'=a+kn$, $b'=ln$, and
$$
  a'b'=(a+kn)(b+ln)=ab+knb+aln+knln = ab+(kb+la+kln)n
$$
from which is it clear that $a'b'\equiv ab\pmod n$, which completes the proof.
A: This is an elementary property of the product in modular arithmetic: 
$$(ab)\bmod c=(a\bmod c)(b\bmod c).$$
(More precisely, $((a\bmod c)(b\bmod c))\bmod c$ as the product can exceed $c$.)

Why is it so ?
$(a+kc)b=ab+(kb)c$: adding a multiple of $c$ to a factor adds a multiple of $c$ to the product, hence nothing changes, modulo $c$.
The same holds for addition or subtraction: $$(a\pm b)\bmod c=(a\bmod c)\pm(b\bmod c).$$
(More precisely, $((a\bmod c)\pm(b\bmod c))\bmod c$ as the sum can exceed $c$ or the difference be negative.)
