TITLE: Explicit expression for recursively defined functions
QUESTION [5 upvotes]: Consider a function $w(i)$, $i \in \mathbb{N}$, defined recursively by:
$w(0)=w(1)=1$, and
$w(i)={i}^{n}-\sum_{j=1}^{i-1}{i \choose j}w(j)$ for  $i>1$.
Is it possible to write $w(i)$ out explicitly as a function of $i$ (i.e., with only $i$ on the right hand side), for $i>1$ ?
In general, given a recursively defined total computable function say, $f(n)$, is it always possible, in principle, to write out $f(n)$ explicitly as a function of $n$, for $n$ large enough? I suspect it's impossible. If so, what is the real reason that prevents us from doing so? 
Another well known example that comes to my mind is:
$f(0)=f(1)=c$, and 
$f(i)=5f(i-1)(1-f(i-1))$ 
where we can choose some value of $c \in (0,1)$ so that the trajectory of $f(i)$ becomes chaotic in $(0,1)$. If we could really express any recursively defined total computable function explicitly, then we could write out the chaotic trajectory in a finite, flat and non-dynamic formula, which seems strange to me. (or can we?) 

Edit: Thank you very much for all the answers. They are really helpful. 
@Joel: Is the recursive formation essential when the resulting recursive function doesn't grow faster than it's component functions ${g}_{i}$ (in your notation)? i.e., is it true that given ${g}_{i}$ and recursive function $f$ defined on them, as long as $f$ doesn't grow faster than the fastest of ${g}_{i}$, the recursive formation can be dispensed with?
@Carl: By "...it is not possible to come up with a completely explicit form for every computable function, in some fixed signature" , did you mean to write out the functions symbolically using finite, fixed set of symbols? In @Joel's interpretation, he didn't distinguish between we being able to symbolically write out exponential functions as, say ${2}^{n}$, and just taking the recursive function on multiplication as another "known" basic function ${g}_{i}$. Of course, one may just dismiss ${2}^{n}$ as a "shorthand" for the whole recursive formation, but I do feel they are not quite the same ("${2}^{n}$" is closer to what I had in mind by "explicit". I'd like to know if the distinction makes any sense in any sense). 
Another question I have concerns about this "...because then by enumerating all possible forms we would get a numbering ϕ as in the theorem" But just because we can write a completely explicit form for each computable function doesn't mean we can enumerate these forms effectively, does it? If we can't, this does not contradict the theorem. Can you be more specific about this? Thanks!

REPLY [5 votes]: To follow up on Joel Hamkins' answer, the fundamental obstruction here is totality. Explicit definitions are always going to give total functions. The reason that recursion is so powerful for defining functions is that it is possible to make partial functions, e.g. 
$$
f(0) := 0 \qquad f(2n+2) := f(2n) \qquad f(2n+1) := f(2n+3)
$$
The following is a standard theorem.

Theorem. Let $C$ be a countable system of total computable functions $\mathbb{N} \to \mathbb{N}$ with the property

There is a numbering $\phi_i\colon \mathbb{N} \to C$ such that every $f \in C$ is of the form $\phi_i$ for at least one $i$, and there is a uniform way to compute $\phi_i(j)$ given just $i$ and $j$. 

Then $C$ does not include all total computable functions from $\mathbb{N}$ to $\mathbb{N}$.
Proof. Diagonalization. 

In the context of this question, the philosophical meaning of this theorem is that it is not possible to come up with a completely explicit form for every computable function, in some fixed signature, because then by enumerating all possible forms we would get a numbering $\phi$ as in the theorem.
The closest we can hope for in terms of an explicit form for all total computable functions is something like Kleene normal form, but this still includes an unbounded search. Kleene's normal form theorem says that there is a primitive recursive function $U$ and a primitive recursive relation $T$ such that for every computable function $f$ there is an $e$ such that for all $i$,
$$
f(i) \simeq U(\mu s . T(e,i,s)).
$$
where $\mu$ is the unbounded search operator.