calculus - If $sum a_n$ converges absolutely, then $sum (e^{a_n} - 1)$ converges absolutely.

Let $a_n$ be a sequence of real numbers (note necessarily positive) such that $\sum a_n$ converges absolutely. Prove that $\sum e^{a_n}-1$ converges absolutely

My attempt: We use Cauchy's criterion. For every $p,q\in\mathbb{N}$, we have that
$$
\left|\sum\limits_{n=p+1}^q|e^{a_n} - 1|\right| = \sum\limits_{n=p+1}^q|e^{a_n}-1|
$$
Using the power series representation of the exponent, we have that
$$

e^{a_n}-1 = \sum\limits_{k=1}^{\infty}\frac{a_n^k}{k!}
$$
and therefore
$$
\sum\limits_{n=p+1}^q|e^{a_n}-1| = \sum\limits_{n=p+1}^q\left|\sum\limits_{k=1}^{\infty}\frac{a_n^k}{k!}\right| = \sum\limits_{n=p+1}^q\left|\lim\limits_{j\to\infty}\sum\limits_{k=1}^{j}\frac{a_n^k}{k!}\right|
$$
Since the absolute value is continuous, we can take the limit out of the sum, and then use the triangle inequality to get
$$
\sum\limits_{n=p+1}^q\lim\limits_{j\to\infty}\left|\sum\limits_{k=1}^{j}\frac{a_n^k}{k!}\right| \leq \sum\limits_{n=p+1}^q\lim\limits_{j\to\infty}\sum\limits_{k=1}^j\frac{|a_n|^k}{k!}
$$

Since the outer sum is finite, we can switch it with the limit and switch summation order to get
$$
= \lim\limits_{j\to\infty}\sum\limits_{k=1}^j\frac{1}{k!}\sum\limits_{n=p+1}^q|a_n|^k
$$
Since $\sum |a_n|$ converges, then from some $N_1$, $|a_n| < 1$, and therefore for all $k$, we will have that $|a_n|^k \leq |a_n|$. Let $p,q > N_1$. Then
$$
\lim\limits_{j\to\infty}\sum\limits_{k=1}^j\frac{1}{k!}\sum\limits_{n=p+1}^q|a_n|^k \leq\lim\limits_{j\to\infty}\sum\limits_{k=1}^j\frac{1}{k!}\sum\limits_{n=p+1}^q|a_n|
$$
Let $\varepsilon > 0$. Since $\sum|a_n|$ converges, then it satisfies Cauchy's criterion, and therefore there exists $N_2$ such that for all $p,q > N_2$,
$$

\sum\limits_{n=p+1}^q|a_n| < \frac{\varepsilon}{e-1}
$$
To finish, we pick $N = \max(N_1,N_2)$, and then for all $p,q>N$, we have
$$
\left|\sum\limits_{n=p+1}^q|e^{a_n} - 1|\right| = \sum\limits_{n=p+1}^q|e^{a_n}-1| \leq \lim\limits_{j\to\infty}\sum\limits_{k=1}^j\frac{1}{k!}\frac{\varepsilon}{e-1} = (e-1)\frac{\varepsilon}{e-1} = \varepsilon
$$
as needed.

How is the proof? Can someone come up with a shorter, less complicated proof for this?

Answer

Your proof is fine, just a bit verbose. As mentioned in the comments, the key inequality is
$$
|e^t-1|\le c|t|\tag{*}
$$
for $|t|\le1$. But you can get to ($*$) using your power series approach as well: If $|t|\le1$, then
$$
|e^{t}-1|=\left|\sum_{k=1}^\infty\frac{t^k}{k!}\right|\le \sum_{k=1}^\infty\frac{|t|^k}{k!}\le|t|\sum_1^\infty\frac1{k!}=(e-1)|t|.$$
Once you've established ($*$), you can continue with the Cauchy criterion: If $N_1$ is such that $|a_n|<1$ whenever $n>N_1$, then if $p,q>N_1$
$$
\sum_{n=p+1}^q|e^{a_n}-1|\le\sum_{p+1}^qc|a_n|,

$$
or just use (*) directly in the Comparison test.