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{
A Combinatorial Proof of $log(e^z)=z$
}
\rm
\bigskip
\centerline{ {\it 
Doron ZEILBERGER
}\footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Rutgers University (New Brunswick),
Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway,
NJ 08854-8019, USA.
%\break
{\eighttt zeilberg at math dot rutgers dot edu} , 
%\hfill \break
{\eighttt http://www.math.rutgers.edu/\Tilde zeilberg/.}
Sept 27, 2005.
Exclusively published, as a class-room note
in
\hfill\break
{\eighttt http://www.math.rutgers.edu/\Tilde zeilberg/mamarim/mamarimPDF/hector.pdf} .
Supported in part by the NSF. 
} 
}

As he was rushing to his Complex Analysis graduate class,
my colleague, Hector Sussmann, asked me whether
I can find a formal-power-series-proof of $log(e^z)=z$, where
$$
e^z:=\sum_{k=0}^{\infty} {{z^k} \over {k!}} \quad,
$$
and
$$
log(1+z):=\sum_{k=1}^{\infty} (-1)^{k-1} {{z^k} \over {k}} \quad.
$$
Of course, Hector was aware that one way is to use
`calculus', of the formal kind
(of course, he knows that I don't believe in any other versions!),
as outlined, for example, in my classroom note

{\tt http://www.math.rutgers.edu/\Tilde zeilberg/mamarim/mamarimPDF/lag.pdf }.

Indeed $(\log(e^z))'=(e^z)'/e^z=e^z/e^z=1$ and `integrating' gives the desired result,
but he asked for a {\it direct} calculus-free proof, even of the {\it good} (i.e. formal)
kind.

Here goes
$$
log(e^z)=log(1+\sum_{k=1}^{\infty} {{z^k} \over {k!}})=
\sum_{k=1}^{\infty} (-1)^{k-1} {{1} \over {k}}
(\sum_{a=1}^{\infty} {{z^a} \over {a!}})^k
$$
$$
=\sum_{k=1}^{\infty} (-1)^{k-1} {{1} \over {k}}
\prod_{i=1}^{k} (\sum_{a_i=1}^{\infty} {{z^a_i} \over {a_i!}})
$$
$$
=\sum_{k=1}^{\infty} (-1)^{k-1} {{1} \over {k}}
\sum_{n=1}^{\infty} 
\left ( \sum_{ {(a_1, \dots , a_k) \in N^k} \atop {a_1+ \dots +a_k=n}}
{{1} \over {a_1! \cdots a_k!}}
\right ) z^n
$$
$$
=\sum_{k=1}^{\infty} (-1)^{k-1} {{1} \over {k}}
\sum_{n=1}^{\infty}   k! S(n,k) {{z^n} \over {n!}} \quad,
$$
where $S(n,k)$, the Stirling numbers of the second kind,
are the number of set partitions of  $\{1,2, \dots, n \}$ into $k$
sets. It is well-known and trivial to see that
$$
S(n,k)=S(n-1,k-1)+kS(n-1,k) \quad,
$$
(element $n$ may be either a loner, in which case the rest are partitioned into $k-1$ sets
or it joins one of the $k$ existing sets).
Continuing, we get, by changing order of summation,
$$
=
\sum_{n=1}^{\infty}  
\left ( \sum_{k=1}^n (-1)^{k-1} (k-1)! S(n,k) \right ) {{z^n} \over {n!}}
$$
The coefficient of $z^n/n!$ is
$$
\sum_{k=1}^{n} (-1)^{k-1} (k-1)! [S(n-1,k-1)+kS(n-1,k)] =
$$
$$
\sum_{k=1}^{n} (-1)^{k-1} (k-1)! S(n-1,k-1)
+
\sum_{k=1}^{n-1} (-1)^{k-1} k! S(n-1,k)
=
$$
$$
\sum_{k=1}^{n-1} (-1)^{k} k! S(n-1,k)
- \sum_{k=1}^{n-1} (-1)^{k} k! S(n-1,k)=0 \quad ,
$$
if $n>1$. Of course, when $n=1$ it is $1$.
\end

