If we assume that every differentiable function can be expressed in a power series:$\sum_{n=0}^{\infty}a_{n}x^{n}$and we know that$\frac{ dy }{ dx }x^n=nx ^{n-1}$Then if we try to find the function f(x) that folowes the rule that f'(x)=f(x) and f(0) =1 we can show that for the power series expansion for that function:$\frac{ d }{ dx }\sum_{n=0}^{\infty}a _{n}x ^{n}=\sum_{n=0}^{\infty}na _{n}x ^{n-1}$And as these should be equal then$a _{n-1}x ^{n-1}= na _{n}x ^{n-1} \rightarrow a _{n-1} =n a _{n}\rightarrow a _{n}=\frac{ a _{n-1} }{ n }$So because a(0) must be 1 so that f(0) can be 1 we get$a_{1}=\frac{ 1 }{ 1 }, a _{2}=\frac{ 1 }{ 1*2 }, a _{3}=\frac{ 1 }{ 1*2*3 },...a _{n}=\frac{ 1 }{ 1*2*3,....,*n }$ So the function were after is$f(x) =\sum_{n=0}^{\infty}\frac{ x^n }{ n! }$Unfortunately this does not tell you what that funcion is, but it still tells you it's the function that is it's own derivative. So if you find a function that is it's own derivative then this is it's powerseries expansion.