User:Scm

DE implies sum of series
Differentiating . $e^x := \displaystyle \sum_{n = 0}^\infty \frac {x^n} {n!}$ we get
 * $\displaystyle \sum_{n = 1}^\infty \frac {x^{n-1}} {(n-1)!}$

which after adjusting the index is the same. Evaluating in $0$ yields one, hence the series solves the DE and by uniqueness has to be equal to $\exp(x)$. -Scm
 * I considered doing that, but I don't know how to adjust indices. Wanna teach me? --GFauxPas 17:06, 9 February 2012 (EST)