Infinite series for $n^x$ Is there a simple infinite sum representation for $a^x$?  I can find $e^x = \displaystyle\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$ and plug in $x = xln(a)$ but I'm wondering if there's something out there that doesn't require the logarithm like that.
 A: The logarithm is not a very big complication, it is just a constant. Maybe this is more clear notation-wise if you consider $a^{x}$ instead of $n^{x}$ or if you use a different letter for the index of the sum.
I think that the short/reasonable answer is no: if you compute the first terms of the Taylor series expansion around $0$ you will see that the constant $\log(a)$ pops up immediately.
A: The question sounds like knowing that $f(x) = \sum a_nx^n$ then asking whether $f(\lambda x)= \sum \lambda^n a_n x^n$ can be represented as another series $\sum b_n x^n$ where $b_n$ somehow do not depend on $\lambda$.
The answer is negative, because of the uniqueness of the power series for an analytic function, which implies that $b_n = \frac{f^{(n)}(0)}{n!} = \lambda^n a_n$.
A: Just like you represent $e^x$ using sum(Taylor polynomial centered at $x=0$) we can represent $\ln(x)$ using sum(this time Taylor polynomial centered $x=1$) to get to the formula:$$a^x=\sum_n\frac{\left(\sum\limits_m \frac{(-1)^m}{m+1}(a-1)^{m+1}\right)^nx^n}{n!}$$
