I noticed relation between modulo operation and number which is power of two
Example I have to calculate $ 3431242341 \bmod 2^5 $, which is $ 5 $ but it is equivalent to $ ( 3431242341 \bmod 2^9 ) \bmod 2^5 $
I tried many examples and it seems to be true in general, and I am not sure if it is a coincidence or true in general that I can use first modulo operation ( greater number) and the result will be the same.
The phenomenon you observed holds in greater generality.
Suppose that $m$ and $n$ are positive integers such that $m$ divides $n$. Then for any integer $a$ we have $(a\bmod n)\bmod m= a\bmod m$.
Certainly $(a\bmod n)\bmod m$ is of the right size, between $0$ and $m-1$.
Since $a$ and $(a\bmod n)$ differ by a multiple of $m$, it follows that the remainder when $(a \bmod n)$ is divided by $m$ is the same as the remainder when $a$ is divided by $m$, which is what we needed to show.
Suppose $a\equiv b \mod pq$ and $b \equiv c \mod p$, then we have $$a=rpq+b=rpq+(sp+c)=(rq+s)p+c$$ so that $a\equiv c \mod p$
$\begin{eqnarray}{\bf Hint}\qquad &&\ (a\ {\rm mod}\ kn)\ {\rm mod}\ n\\ &=&\, (a\ -\ q\:k\:\!\color{#c00}n)\ {\rm mod}\ \color{#c00}n\\ &=&\,\qquad\qquad\ a\,\ {\rm mod}\ n\end{eqnarray}$
Example $ $ The parity of an integer $\,a\,$ is the parity of its least significant (units) decimal digit, i.e.
$\ \ \, \begin{eqnarray} a\ {\rm mod}\ 2\, &=&\, (a\ {\rm mod} &10)& {\rm mod}\ 2\\ &=&\qquad\quad &a_0& {\rm mod}\ 2,\,\ \ a_0\! = \text{units decimal digit of }\, a\end{eqnarray}$
Hence an integer is even iff its units digit is even, and it is divisible by $5$ iff its least digit is, and it is divisible by $10^5$ iff its least digit in radix $10^{\large 9}$ is, $ $since $\ a\ {\rm mod}\ 10^{\large 5} = (a\ {\rm mod}\ 10^{\large 9})\ {\rm mod}\ 10^5.\,$ OP is an analog in radix $\,2\,$ vs. $10.\,$ This is a prototypical example of the method of simpler multiples.
More generally congruences persist mod $\rm\color{#c00}{factors}$ of the modulus, i.e.
$\begin{align} &\bbox[5px,border:1px solid red]{a\equiv \bar a\!\!\!\pmod{\!k\:\!\color{#c00}n}\ \Rightarrow\ a\equiv \bar a\!\!\!\pmod{\!\color{#c00}n}}\\[.4em] \text{by its defining divisibility persists: }&\ \ n\mid kn\mid a-\bar a\,\Rightarrow\, n\mid a-\bar a\ \ \text{by transitivity of 'divides',} \end{align}$
OP is a special case of this persistence, by taking $\,\bar a = (a\bmod kn),\,$ and recalling that
$$\,a\equiv \bar a\!\!\!\pmod{\! n}\ \iff \ a\bmod n = \bar a\bmod n$$
Note on $\!\bmod 0\!:\, $ that integer equations in $\,\Bbb Z\,$ persist as congruences $\!\bmod n\,$ can be viewed as the special case $\,k=0\,$ of above, since a congruence $\!\bmod 0\,$ is quivalent to an integer equality $\,a\equiv b\pmod{\!0}\iff 0\mid a-b\iff a=b,\,$ i.e. $\,\Bbb Z_0 = \Bbb Z\bmod 0\cong \Bbb Z.\,$ see here.
