From your question
$x= a_0 + a_1p+ a_2p^2+...+ a_{e-1}p^{e-1}$
This isn't exactly correct.
It should actually be
$x_i = a_0 + a_1{p_i}+ a_2{p_i}^2+...+ a_{e-1}{p_i}^{e-1}$
This is because you have to first convert the DLP in the group to a DLP in the subgroup.
Let $p-1 = p_1^{e_1}.p_2^{e_2}... p_n^{e_n}$
By Lagrange's Theorem, a cyclic group of order $p-1$ has a cyclic subgroup of corresponding to each of the factors & subfactors of the group.
If order of the group is $N = a*b*c$, then will be cyclic subgroup of order $a$, one of order $b$ & one of $c$.
If $g$ is the generator of the main group, then the generator for the subgroup of order $a$ is $g^{\frac {N} {a}}$.
So the subgroup of order $p_i^{e_i}$ has a generator $g^{\frac {p-1}{p_i^{e_i}}}$
Let the DLP of the main group be
$g^x = h \pmod p$
Let $r = \frac {p-1}{p_i^{e_i}}$
Let's raise both sides of the DLP by $r$
So $({g^x})^r = h^r \pmod p$
Can be rewritten as
$({g^r})^x = h^r \pmod p$
$g^r$ is the generator of the subgroup. Let's call it $g_i$
i.e. $g_i = g^r$.
So
$g_i^x = h^r \pmod p$
Let $h_i = h^r$ & swap the left & right side
$h_i = g_i^x \pmod p$
This is now the DLP in the subgroup, since $g_i$ is the generator of the subgroup of order $p_i^{e_i}$
We can do a further simplification.
Since the order of the subgroup is $p_i^{e_i}$, when we solve for $x$ only in this subgroup, the maximum value of $x$ for the solution of the subgroup DLP can only be $p_i^{e_i}$
We can express this condition as a congruence.
$x = x_i \pmod {p_i^{e_i}}$
It's this $x_i$ which you expand as
$x_i = a_0 + a_1{p_i}+ a_2{p_i}^2+...+ a_{e-1}{p_i}^{e-1}$
And the subgroup DLP is
$h_i = g_i^{x_i} \pmod p$
When you solve the DLP in the subgroup, you get $x_i$ & not $x$.
So you have to solve the DLP in each of the $n$ subgroups to get
$x = x_1 \pmod {p_1^{e_1}}, x = x_2 \pmod {p_2^{e_2 }}, ..., x = x_n \pmod {p_n^{e_n}}$
Then you combine all of them with the Chinese Remainder Theorem to get $x$.
Hence to use your own words, you have to "exhaust all factors".
