TITLE: How to prove $e^{\log(1+x)}= 1+x$ by series expansion?
QUESTION [8 upvotes]: as the title says, i want to prove $e^{\log(1+x)}= 1+x$, by substitute $\log(1+x) = \sum _{i=1} ^{\infty} \frac{(-1)^{i+1}x^i}{i}$ and $e^x=\sum _{i=0} ^{\infty} \frac{x^i}{i!} $. Can some one help me?

REPLY [5 votes]: You have:
$$e^{\log (1+x)}=\sum_{k=0}^\infty \frac{1}{k!}\left(\sum_{m=1}^\infty \frac{(-1)^{m+1} x^{m}}{m}\right)^k=\sum_{n=0}^\infty a_n x^n$$
The zero order can only be achieved when $k=0$, so that $a_0 = 1$. As for $a_1$, only $m=k = 1$ contributes, so again $a_1 = 1$. 
But how do we prove all higher orders are zero?  
If you take the term by term derivative, you should get: 
$$\sum _{k=0}^{\infty } \frac{k \left(\sum _{m=1}^{\infty } (-1)^{m+1}
   x^{m-1}\right) \left(\sum _{m=1}^{\infty } (-1)^{m+1}
   x^m / m\right){}^{k-1}}{k!} =$$
$$\sum _{k=0}^{\infty } \frac{1}{(k-1)!}\cdot\frac{1}{1+x}\cdot \log^{k-1}(1+x) = \frac{e^{\log (1+x)}}{1+x}$$
From here one has:
$$\sum a_n n (x +1)x^{n-1} = \sum a_n x^n$$
Which can be rewriten as:
$$-a_0 + a_1 + 2 a_2 x + \sum_{n=2}^\infty ((n - 1)a_n + (n+1)a_{n+1})x^n = 0$$
Since this must hold true for all $x$, it is clear that for all $n>1$, $a_n=0$.