Find the Maclaurin Series for $e^{\arctan x}$ up to $\mathcal{O}(n^5)$ The question is to find the value of the Maclaurin series expansion of $e^{\arctan x}$ up to (but not including)  $\mathcal{O}(n^5)$.
I tried using the Maclaurin series for $e^{u}$ then subbing $\arctan x$ into it, and I got  $$\sum_{n=0}^{\infty } \frac{(\arctan x)^{n}}{n!}$$
Is this the correct approach to solving this problem or am I missing something? Cause if so I'm not sure when to stop to reach $\mathcal{O}(n^5)$

https://en.wikipedia.org/wiki/Exponential_formula
\begin{align}
\arctan x & = x - \frac{x^3} 3 + \frac{x^5} 5 - \frac{x^7} 7 + \cdots \\[10pt]
& = a_1 x + a_2 \frac{x^2} 2 + a_3 \frac{x^3} 6 + a_4 \frac{x^4}{24} + \cdots \\[10pt]
a_1 & = 1 \\
a_2 & = 0 \\
a_3 & = -1/3 \\
a_4 & = 0 \\
a_5 & = +1/5 \\
& \,\,\,\vdots \\[10pt]
e^{\arctan x} & = 1 + b_1 x + b_2 \frac{x^2} 2 + b_3 \frac{x^3} 6 + b_4 \frac{x^4} {24} + \cdots \\[10pt]
b_1 & = a_1 \\
b_2 & = a_1^2 + a_2 \\
b_3 & = a_1^3 + 3a_1 a_2 + a_3 \\
b_4 & = a_1^4 + 4a_1 a_3 + 3a_2^2 + 6a_1^2 a_2 + a_4 \\
& \text{etc.}
\end{align}
The pattern is this:
The coefficient of $a_2^2$ is $3$ because there are three ways to partition a set of four objects into two sets of two:
$$
ab/cd, \qquad ac/bd, \qquad ad/bc
$$
The coefficient of $a_1^2 a_2$ is $6$ because there are six ways to partition a set of four objects into two sets of one and a set of two:
$$
a/b/cd, \qquad a/c/bd, \qquad a/d/bc, \qquad b/c/ad, \qquad b/d/ac, \qquad c/d/ab
$$
And so on.
Thus
$$
e^{\arctan x} = 1 + x + x^2 + \frac{x^3} 9 - \frac{x^4}{72} + \cdots
$$
You should probably check my arithmetic.
Next we have
$$
b_5 = a_1^5 + 10a_1^3 a_2 + 15a_1 a^2_2 + 10a_1^2 a_3 + 10a_2 a_3 + 5a_1 a_4 + a_5.
$$
