Maximum integer $n$ satisfying Taylor approximation 
What is the maximum $n \in \mathbb N$ satisfying the formula
  $$\sinh(x^4) = x^4 + \frac 1 6 x^{12} + \mathcal o(x^n) \quad \mathrm{for}\ x\to0 $$

The Taylor's theorem assures me that
$$\sinh x = x + \frac{x^3}{3!} + o(x^3)\ \mathrm{or}\ o(x^4) \quad \mathrm{for}\ x \to 0$$
which should mean
$$\sinh x^4 = x^4 + \frac{x^{12}}{3!} + o(x^{12})\ \mathrm{or}\ o(x^{16}) \quad \mathrm{for}\ x \to 0$$
However, the answer seems to be $n= 19$ rather than $16$. What am I missing? Is there a faster and more rigorous way to do these?

As you know the taylor series of the $\sinh(x)$ is:
$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!} = x + \frac{x^3}{3!} + \frac{x^5}{5!}+\cdots$$
Therfore, 
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + \frac{x^{20}}{5!}+\cdots$$
Hence for integer powers of $x$:
$$\sinh(x^4) = x^4 + \frac{x^{12}}{3!} + O(x^{20}) = x^4 + \frac{x^{12}}{3!} + o(x^{19})$$
If you know the difference of big-$O$ and small-$o$ notation the above equality makes sense for you when $x\rightarrow 0$.
