Find a positive integer $n$ such that its digits product is $n^2 -13n-25$ 
Find a positive integer $n$ such that its digits product is $n^2 -13n-25$   


My attempt
If we assume that $n$ is a two digit integer then let $n=10k+r$ where $1 \leq k \leq 9 $ and $0 \leq r \leq 9$ . Plugging $n=10k+r$ in $n^2 -13n-25$  we get $kr=(10k+r)^2 -13(10k+r)-25$. Manipulating this expression as a quadratic for $r$ we conclude that $n=15$ is a solution.  
My question
I can't generalize it for any $n$ greater than two digit. Can anyone help? Thanks in advance.
 A: Notice that the digital product of $n$ is at most $9^m$, where $m$ is the number of digits of $n$. Then, for all values of $n$ with three digits or more,
$$9^m \le 9^{\log_{10}(n)+1} < 10^{\log_{10}(n)+1} = 10n < n^2 - 13n - 25.$$
Hence any suitable value of $n$ must have two digits or fewer.
A: assume that n is  $\ m \ $  digit integer 
$ n=\sum_{i=0}^{m-1}a_{i}10^i \\
\prod_{i=0}^{m-1}a_{i}=n^2-13n-25 \geqslant 0 \Rightarrow \boxed{15\leqslant n} \\
\prod_{i=0}^{m-1}a_{i}=a_{0}\times a_{1}\times a_{2}\times ....\times a_{m-2}\times a_{m-1}\leqslant 9^{m-1}\times a_{m-1}< 10^{m-1}\times a_{m-1}\leqslant n  \\
n^2-13n-25\leqslant n \\
n^2-14n-25\leqslant 0  \Rightarrow \boxed{n< 16} \\$ 
Hence $\boxed{n=15}$
A: Let us assume that $n>10^{20}$. Realize that no digit of $n$ can be $0$. Also, realize that $n^2-13n-25>n$. 
Then realize that $n^2-3n-25>n>9^{\text{number of digits of n}} \ge \text{digit product of } n$ 
So we conclude that no such $n$ can be more than $21$ digits. 
EDIT
A far tighter bound can be achieved, as pointed out by @JoshuaCiappara. 
If $n$ is $3$ digits, $$n^2-13n-25 \ge 100^2-13 \times 100 -25 >729 \ge \text{digit product of } n$$
The same is true when $n$ is $4$ digits or higher. 
So $n$ can be at most $2$ digits. Thus the answer is $15$. 
