Motivated by [this question](https://mathoverflow.net/questions/21290/whats-an-example-of-a-transendental-power-series), I remembered a question I was curious about sometime which I am sure has some easy and nice example for it as well, which I just can't think of for some reason. I want an example of a power series that is not differentially algebraic. A differential algebraic power series is a series $f(t)$ satisfying an equation $P(t,f(t),f'(t),\ldots,f^{(k)}(t))=0$ for some $k$ and some polynomial $P$ in $k+2$ variables. 


**Update:** examples in the comments below ($\sum t^{n^n}$, $\sum t^{2^n}$) make me ask a refinement (of a sort) for the original question: these examples are reminiscent of all those Liouville-flavoured examples of transcendental numbers, - I wonder if there is a Liouville-flavoured proof, stating that if the polynomial P is of given (multi)degree, some inequality holds that is obviously impossible for the series above?


**Update 2:** there are quite a few examples now, and I am tempted to accept the $\sum t^{n^n}$ answer since the example itself is easy and it came together with an easy explanation. I wonder what are other general approaches besides the ones that are exhibited in answers here (looking at p-adic norms of coefficients and looking at powers of $t$ with nonzero coefficients).



An example of a series that is not differentially algebraic? 
You'd be better off in characteristic zero, for $f^{(p)}(t)=0$ in characteristic $p$. Then the sea of zeroes example $\sum\_n t^{n^n}$ will do the trick. For large enough $n$, there will be a cluster of non-zeroes in degrees $kn^n-m$ for small (and bounded) $k$ and $m$, "reachable" only by products of $(t^{n^n})^{(s)}$, the same $n$, bounded $s$. Their vanishing will give infinitely many linear relations on the coefficients of $P$, which you can explicitly write down and see that there are no nonzero solutions on coefficients.


