<s>[INST] How does one show that multiples of number does not have a zero in the decimal expansion? 
Prove that there exists a number divisible by $5^{1000}$ not containing a single zero in its decimal notation.

The above question is taken from this site. It is question no. 88 in the list. I could not find any solutions for this problem. Apparently a "schoolboy" should be able to solve this question.
In general, I have a hunch that this statement must be true for any number (instead of $5^{1000}$). However, I cannot think of a strategy to show there are no zeros in the decimal expansion. Trailing zeros are trivial to catch but zeros in the middle confound me.
Is there a clever construction for the case of $5^{1000}$?
Edit: As explained in the comments, my hunch does not make sense for a multiple of 10. 
 [/INST] Idea of Approach
By induction, an using the fact that $5^{n}$ is a divisor of $10^n$ and $10^{n-1}$ leaves only a certain kind of remainder when divided by $5^n$.
Claim and proof
Claim : For all $n$, there exists an $n$ digit multiple of $5^n$ that does not contain any zeros.
Proof : Start with the base case : $5 , 25  , 125$ for $n = 1,2,3$. 
The inductive case : let $L$ be an $n$ digit number that has no non-zero digits and is a multiple of $5^n$. We want a number of the form $a \times 10^{n} + L$(this will have $n+1$ digits) that divides $5^{n+1}$, where $a \neq 0$.
Well, if $a \times 10^n + L$ is a multiple of $5^{n+1}$ then the remainders when you divide $a \times 10^n$ and $L$ by $5^{n+1}$ respectively, must sum up to a multiple of $5^{n+1}$.
But both are multiples of $5^n$, so $10^n$ leaves a remainder of $b 5^n$ when divided by $5^{n+1}$ and $L$ leaves a remainder of $c 5^n$ when divided by $5^{n+1}$. 
Hence, $a 10^n + L$ leaves the remainder $(ab+c)5^n$ when divided by $5^{n+1}$.
Note that $b$ cannot be zero since $10^n$ is not a multiple of $5^{n+1}$. But $c = 0$ is possible.
So the question is : can we make $ab+c$ a multiple of $5$ everytime?
If $c = 0$  then take $a = 5$, since in that case, regardless of $b$ we have $ab+c$ is a multiple of $5$.
If $c \neq 0$ then since $b \neq 0$ we take $a = (-c)b^{-1} \mod 5$, where $b^{-1}$ is the multiplicative inverse of $b$ modulo $5$. 
Therefore, in either case, we are done.
Example


*

*Let us continue from $125$. So we have to add $a \times 10^3 + 125$ for some $a$. Here, note that $1000$ leaves the remainder $3 \times 125$ when divided by $5^4 = 625$ so $b = 3$ and $L$ leaves the remainder $1 \times 125$ when divided by $625$, so $c = 1$. So $b^{-1} = 2$(since $2 \times 3 = 1 \mod 5$) and $b^{-1} \times -c = -2  = 3 \mod 5$. Check that $3125$ is a four digit multiple of $625$.

*Next, we want a five digit multiple of $5^5 = 3125$. Here, we have $c = 0$ because $L = 3125$ is a multiple (in fact equal) to $3125$, so we have $a = 5$. Check that $53125$ is a multiple of $3125$.

*Next, we want a six digit multiple of $5^6 = 15625$. Here, check that $L =53125$ ,  $c = 2$ and $ b= 2$ as well, so $b^{-1} = 3$ and $3 \times -2 = -6 \equiv 4$ so $453125$ is a multiple of $15625$.
Conclusion
Hence, for all $n$, there exists a multiple of $5^n$ which has exactly $n$ digits, and furthermore consists only of the digits $1,2,3,4,5$ by construction.
Addendum
Maybe the approach might seem a little arbitrary. I actually got the idea from another question I had solved about eight and a half years back (and wrote down the first $21$ terms on the school board!) : Show that there is a multiple of $2^n$ that consists only of the digits $6$ and $7$.
The style of proof there was exactly the same : take such a multiple of $2^n$ and adjoin either a $6$ or $7$ on the left so that things work out for the next power.
Last but not the least, note that the post attached in the comments to the question addresses this question but does not give a proof. Rather, it uses this fact to prove that every number not ending with zero has a multiple with no zero in its representation. In fact, as pointed out, such a number can also be made to consist solely of the digits $1,2,3,4,5,6,7$.
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