How do I show that formal logarithm is the inverse of the formal exponential? Let $A$ be a unital commutative and associative $\mathbb{Q}$-algebra.
Define $exp(f):=\sum_{n=0}^\infty \frac{f^n}{n!}$ for each $f\in XA[[X]]$.
Define $log(f):=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (f-1)^n$ for each $f\in 1+XA[[X]]$.
Hence, we have two maps $exp:XA[[X]]\rightarrow 1+XA[[X]]$ and $log:1+XA[[X]]\rightarrow XA[[X]]$.
I am trying to prove that $log$ map is the inverse of the $exp$ map.
The first thing I tried is to directly show $log\circ exp=id$ and $exp\circ log = id$ by checking if the identities hold for every element $f$, but this does not work well since this way involves too many calculations. For example, $$[X^n]exp(log(f))=[X^n]\sum_{k=0}^n log(f)^k/k! = \sum_{k=0}^n \frac{1}{k!} [X^n]log(f)^k= \sum_{k=0}^n \frac{1}{k!} [X^n](\sum_{l=1}^k \frac{(-1)^{l+1}}{l} (f-1)^l)^k$$.
Thus, we have to show that $$\sum_{k=0}^n \frac{1}{k!} [X^n](\sum_{l=1}^k \frac{(-1)^{l+1}}{l} (f-1)^l)^k=[X^n]f$$.
But this calculation is really a nightmare.. Is there a clever way to show this? If not, how do I wisely calculate to show the above identity?
Thank you in advance.

Another way might be
(feeling too lazy right now
to work out
all the details)
to use the functional equations:
For ln,
$f(xy) = f(x)+f(y),
f'(1) = 1$.
For exp,
$g(x+y) = g(x)g(y),
g'(0) = 1$.
From these,
can get
$f'(x)
=\dfrac1{x}
$
and
$g'(x) = g(x)$.
