Let's prove it with an induction for fun
Call an integer $n-good$ if it has the property for an integer n
for $n=1$ simply take $20221$
Suppose there's a positive integer $k_{n-1}$ with $gcd(k_{n-1},10)=1$ that's $(n-1)-good$ for a positive integer $n$
for $n$, let $n=2^l5^kn_0$, where $gcd(10,n_0)=1$ and take
$$k_{n}=10^xa+k_{n-1}$$where $x$ is a big enough integer and $a=2^k5^la_0$, where $a_0\equiv20221(n_0k_{n-1}^{n-1})^{-1}(mod\ 10^5)$
For $1\le i\le n-1$, as $x$ is big enough, $(10^xa+k_{n-1})^i$ will have $k_{n-1}^i$ in its decimal representation, therefore, a $2022$-block in it.
For $i=n$, we'll have that $(10^xa+k_{n-1})^n$ will have
$$10^x\binom n1ak_{n-1}^{n-1}=10^{x+k+l}n_0a_0k_{n-1}^{n-1}$$in its representation, but
$$a_0n_0k_{n-1}^{n-1}\equiv 20221(mod\ 10^5)$$a $2022$-block, as we wanted.