Looking at even indices, from $a_{2j_0+2} \approx \frac{1}{j} a_{2j_0}$, we get $a_{2j_0+2k} \approx \frac{(j_0-1)!}{(j_0+k-1)!}a_{2j_0} = \frac{1}{(j_0+k)!} C$: $$a_{2j_0+2k} \approx \frac{(j_0-1)!}{(j_0+k-1)!}a_{2j_0} = \frac{1}{(j_0+k)!} C \tag{1}$$, with $C=(j_0+k)\,(j_0-1)! \,\,a_{2j_0}$
So, taking $n= 2(j_0+k)$, we have finally $a_{n} \approx \frac{1}{(n/2!)} \, C$
Finally, $h_{even}(\xi) \approx \sum\limits_{n= j_0 + 1}^{+\infty}a_{2n} \xi^{2n}= C\sum\limits_{n= j_0 + 1}^{+\infty}\frac{1}{n!} (\xi^2)^n $
The behaviour of $h_{even}(\xi)$ for high $\xi$, is then the same as the behaviour of $C\sum\limits_{n= 0}^{+\infty}\frac{1}{n!} (\xi^2)^n = C e^{\xi^2}$
You have the same demonstration for $h_{odd}(\xi)$, so finally, the behaviour of $h(\xi)$ for high $\xi$ is $C e^{\xi^2}$, too.
UPDATE (to answer to the OP's comments)
a) You have : $a_{2j_0+2} \approx \frac{1}{j} a_{2j_0}$, then you apply the formula to $(j_0+k)$ : $a_{2j_0+2k} \approx \frac{1}{j_0+k-1} a_{2j_0+2k-2}$, so finally, you have : $a_{2j_0+2k} = \frac{1}{(j_0+k-1)(j_0+k-2)...(j_0)}a_{2j_0}$. You multiply numerator and denominator by $(j_0+k)(j_0-1)!$, and you get the formula $(1)$
b) The identity is $e^y = \sum\limits_{i=0}^{+\infty}\dfrac{y^n}{n!}$
c) You have to choose some particular $j_0$ as a departure point for the recurrence relation described in a). I used this notation to show that $j_0$ is not a variable, it is fixed.
d) The identity described in b) is true for all $y$, because the serie is convergent for all $y$