Consider $2^n\pmod{10^{k+1}}$. As $2\equiv 2\pmod{10^{k+1}}$, and thus there exists $n$ such that $2^n$ ends with $2$ at the end and block of $m\ge k\ 0$s. If $m=k$ we are done. If not, take $2\cdot 2^n, 2^2\cdot 2^n\cdots$. Some power of $2$ among this sequence has the desired number of $0$s.

iarnab_kundu wrote: Consider $2^n\pmod{10^{k+1}}$. As $2\equiv 2\pmod{10^{k+1}}$, and thus there exists $n$ such that $2^n$ ends with $2$ at the end and block of $m\ge k\ 0$s $\dots$ Unfortunately this is false, since $2$ and $10$ are not relatively prime. As an illustration, consider the following: There is no number (with $3$ or more digits) which is simultaneously a multiple of $4$ and ends on $\dots 02$. This is because $100$ is divisible by $4$, and hence this number is $\equiv 2\pmod{4}$.