convergence in $\hat{\mathbb{Z}}$, modulo prime power The following problem appears in Lenstra's Galois Theory for Schemes (p 14, Ex 1.16).

Let $b\in\mathbb Z_{\ge0}$. Define the sequence $(a_n)_{n=0}^\infty$
  by $a_0=b, a_{n+1}=2^{a_n}$. Prove that $(a_n)_{n=0}^\infty$
  converges in $\hat{\mathbb Z}$, and that $a:=\lim_{n\to\infty}a_n$ is
  independent of $b$.
Then, write $a=\sum_{n=1}^\infty c_n n!$. Compute $c_n$ for
  $n=1,\cdots, 10$.

My first question is about finding $c_1, \cdots, c_{10}$. Noting the fact that $ \hat{\mathbb Z} \simeq\varprojlim\mathbb Z/n!\mathbb Z$, I have written down a bijection from $\varprojlim\mathbb Z/n!\mathbb Z$ onto the collection $\{\sum_{n=1}^\infty c_n n!: 0\leq c_n\leq n\}$ of formal series. According to the bijections I need to find out $a\text{ mod } n!$ for $n=2,\cdots, 11$. Thanks to the Chinese remainder theorem, the problem reduces to the computation of $a$ modulo prime-power factors in each $n!$. With the help of Euler's formula, $a\text{ mod }p$ can be computed for every odd prime $p$; and from the definition of $a$ we have $a\equiv 0\pmod{2^k}$. But I do not have any idea about calculating $a\text{ mod }p^k$. In particular, what I need are $a\text{ mod }3^2$, $a\text{ mod }3^4$ and $a\text{ mod }5^2$.
Then, we turn back to a more fundamental question: what does it mean by the sequence to converge? (Lenstra has not defined convergence in his notes.) Intuitively, $(a_n)_{n=1}^\infty$ converges if for every $M\in\mathbb N$, there is an $N\in\mathbb N$ such that for each $m\leq M$, the expression $a_n\text{ mod }m!$ is a constant for all $n\geq N$. Is this "definition" concise enough? And why do $a_n\text{ mod }{m!}$ become stable eventually? Finally, why the limit is independent of $b$?
 A: Convergence follows from the following statement: if $n\geq N$, then $a_n\pmod{N}$ is independent of $b$. We can prove this by induction on $N$. 
For $N=1$ the statement is trivial. Let $N>1$ and assume that the statement is true up to $N-1$. Let us look at $a_n\pmod{N}$. Writing $N=2^mN'$ where $N'$ is odd, it suffices to show that $a_n\pmod{2^m}$ and $a_n\pmod{N'}$ are independent of $b$, assuming of course $n\geq N$. The condition implies that $n\geq m$, whence $a_n\equiv 0\pmod{2^m}$. The condition also implies that 
$$n-1\geq N-1\geq\varphi(N)\geq\varphi(N'),$$ 
hence $a_{n-1}\pmod{\varphi(N')}$ is independent of $b$ by the induction hypothesis. By the Euler-Fermat theorem, this implies that $a_n\pmod{N'}$ is independent of $b$, and we are done.
Remark. The notion of convergence in $\hat{\mathbb{Z}}$ is given by the profinite topology of $\hat{\mathbb{Z}}$. That is, a sequence converges in $\hat{\mathbb{Z}}$ if and only if, for any $N$, the reduction of the sequence modulo $N$ converges in $\mathbb{Z}/N\mathbb{Z}$. The finite residue ring $\mathbb{Z}/N\mathbb{Z}$ is endowed with the discrete topology, so convergence really means stabilizing modulo $N$, for any given $N$.
