Numbers such that they equal the product of their own digits

Your conjecture is true. If it isn't, let us suppose that there is a number $m>9$ of $n$ digits such that $dp(m)=m$. As you have checked and also proved by @ChristopherMarley in the comments, $n>2$.

Let's chop off the last digit of $m$ to get $m'$ of $n-1$ digits. Now, $m' \leq m/10$ (since we're taking the floor of $m/10$) and $dp(m') \geq m/9$(since the last digit is $9$ at most) $> m/10$.

Hence, $m' \leq dp(m')$. Continuing this chopping off of last digit, we'll get a number $m_0$ of $2$ digits such that $m_0 \leq dp(m_0)$ but we know that's not possible.

So, such an $m$ doesn't exist i.e. your conjecture is true that single digit numbers are the only numbers with $dp(m)=m$.

As you've observed, it's not hard to see that there is no such number with more than one digit. Considering three-digit numbers, just for example, we see that for a number of the form abc to have the desired property, we would need $a\cdot b\cdot c = 100a+10b+c$. The left side is at most $81a$ and the right side is at least $100a$, so they can't be equal. A similar argument works for any numerals of more than a single digit.

But this argument depends on the fact that no digit exceeds 9. If we are willing to relax the rules about base-10 numerals a bit, and allow digits larger than 9, we can find solutions. So for example when we count “seventy-seven, seventy-eight, seventy-nine” instead of continuing with eighty we are allowed to continue “seventy-ten, seventy-eleven, seventy-twelve…”, which we will write 7 7, 7 8, 7 9, 7 10, 7 11, 7 12…

What does a number like “seventy-twelve” mean, anyway? Simple: Just as “seventy-nine” means seven tens and nine units, or $7·10+9 = 79$, “seventy-twelve” means seven tens and twelve units, or $7·10+12 = 82$. 7 12 is just another way to write 8 2. They represent the same number, similar to how $\frac12$ and $\frac48$ are also two different ways to write the same number. In the conventional decimal system there is only one way to write each number. But in this relaxed system, there may be several ways. For example, the number $121$ can be written as 1 2 1 or as 12 1 or as 11 11 or even as 9 31. (That last is nine tens and 31 units, or $9\cdot 10 + 31$.)

Now there are some numbers that are equal to their digital product. For example, 6 12 (“sixty-twelve”) is another way to write $72$, and indeed $6\cdot 12=72$!

Simple algebra ($10a+b = ab$, so we want $a = \frac b{b-10}$) finds the two-digit examples:

\begin{array}{rlr} \text{Relaxed numeral} & \text{Silly name} & \text{Conventional numeral} \\ 11\ 11 & \text{(eleventy-eleven)} & 121\\ 6\ 12 & \text{(sixty-twelve)} & 72 \\ 3\ 15 & \text{(thirty-fifteen)} & 45 \\ 2\ 20 & \text{(twenty-twenty)} & 40 \end{array}

And of course there are solutions with more digits, such as 6 16 8 (six hundred sixteenty-eight) $= 768$.

Note: $dp(N)$ is an integer. So if $dp(N) = N$ then $N$ is an integer.

If $N = \sum\limits_{k=0}^n d_k10^k$ has more than one digit, i.e. $n \ge 1$ and $N$ is such a number then

$d_0d_1.....d_{n-1}d_n =\sum\limits_{k=0}^n d_k10^k > d_n 10^n$

$d_0d_1.....d_{n-1} > 10^n$

$9^n \ge d_0d_1.....d_{n-1} > 10^n$