Is there any deep reason that 23456789 is prime?

This is not an answer, but in base $n$ that number is $$ a_n=\frac{2n^{n-1}-n^{n-2}-n^2+n-1}{(n-1)^2}$$ Which is prime for $n=3,4,6,10,16,18,36$ and no other $n\leq 500$. $\sum_{n=3}^\infty \frac 1{\ln(a_n)}$ converges, so by a heuristic following from the prime number theorem this sequence probably has only finitely many primes.

COMMENT.-Here something I like and and maybe related to the problem proposed (note that with this definition the coefficient $a$ can be extended to be non-digit).

Define

$$P_n(x)=\sum_{2\le a\le n} ax^{n-a}$$

The polynomials $P_n(x)$ have degree $n-2$ and can be defined by recurrence via
$$P_{n+1}(x)=xP_n(x)+(n+1);\space P_2(x)=2$$ This way one has

$$P_9(10)=\color{red}{23456789}\\P_9(4)=50969,\text{ prime }\\P_9(2)=757,\text{ prime}$$ Besides with exception of $P_6$ for which $P_6(n)$ is composite for $1\le n\le 10$ one has the following primes for $1\le n\le 10$ $$\begin{cases}P_8(5)=43943\\P_8(3)=2729\\P_7(3)=907\\P_5(2)=41\\P_5(6)=569\\P_4(3)=31\\P_4(9)=193\\P_3(1)=5\\P_3(2)=7\\P_3(4)=11\\P_3(5)=13\\P_3(7)=17\\P_3(8)=19\\P_3(10)=23\\P_2(n)=2\text{ for all } n\end{cases}$$