Solving an infinite system of equations for the coefficients of a power series for $f(x+1)=\exp(f(x))$ Consider the sequence of formulae, $a_n$, such that $a_0=1$ and for all $n>0$, we have
$$a_n = \sum_{k=1}^nT(n,k)c_k a_{n-k},$$
where $$T(n,k)=\frac{(n-1)!k}{(n-k)!},$$
$c_0=0$, and $c_k$ for $k>0$ are unknown coefficients of a power series (see the background below for more information). Pairing this with the secondary fact that
$$a_n=\sum_{k=n}^\infty \frac{k!}{(k-n)!}c_k,$$

can we solve for all $c_k$?

I am reasonably confident that this is sufficient information (given the background of the problem, as described below, assuming the uniqueness of an analytic solution). However, this effectively surmounts to solving an infinite set of nonlinear equations
$$\sum_{k=n}^\infty \frac{k!}{(k-n)!}c_k - \sum_{k=1}^nT(n,k)c_k a_{n-k} = 0,$$
and I am unaware of any methods which can do this (despite my confidence that this is possible), so any insight or advice would be great.

Background:
This essentially is a continuation of my earlier question (which lead me to a new attack vector as described above), in which we are constructing an analytic function $f:\mathbb{R}\to\mathbb{R}$ where
$$f(0)=0,f(x+1)=\exp(f(x)),\text{ and } x\in[0,1)\implies f(x)=\phi(x),$$
such that,
$$\phi(x)=\sum_{n=0}^\infty c_n x^n.$$
In fact, $a_n=\left(\frac{d}{dx}\right)^n f(1)$. Doing so would allow for (among other things) the explicit construction of an analytic function $h$ such that $h(h(x))=\exp(x)$, as described in this answer.

As requested in the comments, here is a table of some values for $T$. More terms are available on the OEIS as sequence A121757. (Be aware that my definition of $T$, and the respective definition in the OEIS have offset indices by $1$).
$$\begin{array} {|r|r|}\hline n & T(n,1) & T(n,2) & T(n,3) & T(n,4) & T(n,5) & T(n,6) & T(n,7) & T(n,8) & T(n,9) \\ \hline 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 2 & 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 3 & 1 & 4 & 6 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 4 & 1 & 6 & 18 & 24 & 0 & 0 & 0 & 0 & 0 \\ \hline 5 & 1 & 8 & 36 & 96 & 120 & 0 & 0 & 0 & 0 \\ \hline 6 & 1 & 10 & 60 & 240 & 600 & 720 & 0 & 0 & 0 \\ \hline 7 & 1 & 12 & 90 & 480 & 1800 & 4320 & 5040 & 0 & 0 \\ \hline 8 & 1 & 14 & 126 & 840 & 4200 & 15120 & 35280 & 40320 & 0 \\ \hline 9 & 1 & 16 & 168 & 1344 & 8400 & 40320 & 141120 & 322560 & 362880 \\ \hline  \end{array}$$
 A: ----  This is just an addendum to the excellent answer above, concerning  indexing the matrix and vectors from $0$  ---
We can index the matrix and the vector from $0$ putting:
$$
T_{n,m}  = \left( \begin{array}{c}
 n - 1 \\  n - m \\ 
 \end{array} \right)m!\;\,\left| {\;0 \le n,m} \right.
$$
wherefrom for $n=0$ we get
$$
\begin{array}{l}
 a_n  = \sum\limits_{k = 0}^n {T_{n,k} c_k a_{n - k} }
  = \sum\limits_{0 \le k} {T_{n,k} c_k a_{n - k} } \quad  \Rightarrow  \\ 
  \Rightarrow \quad a_0  = T_{0,0} c_0 a_0  = \left( \begin{array}{c}
  - 1 \\  0 \\ 
 \end{array} \right)c_0 a_0  \Rightarrow \left\{ \begin{array}{l}
 c_0  = 1 \\  a_0  = 1 \\ 
 \end{array} \right. \\ 
 a_n  = \sum\limits_{k = n}^\infty  {\frac{{k!}}{{\left( {k - n} \right)!}}c_k } \quad  \Rightarrow \quad a_n
  = n!\sum\limits_{0 \le k} {\left( \begin{array}{l}
 k \\  n \\ 
 \end{array} \right)c_k } \quad  \Rightarrow  \\ 
  \Rightarrow \quad a_0  = \sum\limits_{k = 0}^\infty  {c_k }  \Rightarrow \left\{ \begin{array}{l}
 c_0  = a_0  = 1 \\ 
 \sum\limits_{k = 1}^\infty  {c_k }  = 0 \\ 
 \end{array} \right. \\ 
 \end{array}
$$
