Proof.
We begin with a reminder of the
Binomial Theorem, which we will use shortly.
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
We then consider nth term of the sequence, with \( n \in \mathbb{Z}^+ \), and then apply the Binomial Theorem.
\[
\left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^{n} \binom{n}{k} \left( 1 \right)^{n-k} \left( \frac{x}{n} \right)^k = \sum_{k=0}^{n} \binom{n}{k} \frac{x^k}{n^k} = 1 + \sum_{k=1}^{n} \frac{n!}{k!(n-k)!} \frac{x^k}{n^k}
\]
Note that we bring the \( 0 \)th term out of the sum (this step will make subsequent steps much simpler). We now express \(n! \) and \( n^k \) as \( n! = \left( \prod_{i=0}^{k-1}\text{ }{n-i}\right)(n-k)! \) and \( n^k = \prod_{i=0}^{k-1}\text{ }n \). Our relation then becomes:
\[
\left( 1 + \frac{x}{n} \right)^n = 1 + \sum_{k=1}^{n} \frac{\left( \prod_{i=0}^{k-1}\text{ }{n-i}\right)(n-k)!}{\prod_{i=0}^{k-1}\text{ }n} \frac{x^k}{k!} = 1 + \sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \text{ } 1 - \frac{i}{n} \right) \frac{x^k}{k!}
\]
The next step in the proof is to show that every \( n+1 \)th term is greater than every \( n\)th term. We express the \( n+1\)th term as:
\[
\begin{aligned}
\left( 1 + \frac{x}{n+1} \right)^{n+1} &= 1 + \sum_{k=1}^{n+1} \left( \prod_{i=0}^{k-1} \text{ } 1 - \frac{i}{n+1} \right) \frac{x^k}{k!} \\
&= 1 + \sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \text{ } 1 - \frac{i}{n+1} \right) \frac{x^k}{k!} + \prod_{i=0}^{n} \text{ } \left( 1 - \frac{i}{n+1} \right) \frac{x^{n+1}}{(n+1)!} \\
\end{aligned}
\]
We now we have to do some index gymnastics to show that the \( n+1\)th term is greater than the \( n\)th term. Let \( 1 \leq k \leq n \) and \( 0 \leq i \leq k-1 \). We then have
\[
n < n+1 \to \frac{1}{n} > \frac{1}{n+1} \to 1 - \frac{i}{n} < 1 - \frac{i}{n+1} \to 1 - \frac{i}{n} < 1 - \frac{i}{n+1}
\]
Furthermore, we have \( 0 \leq i \leq k-1 \leq k \leq n \), which implies that \( 0 \leq i \leq n \). We then have
\[
0 \leq 0 \leq n \to 0 \leq \frac{i}{n} \leq 1 \to 1 - \frac{i}{n} > 0
\]
We have shown that \( 1- \frac{i}{n} \leq 1 - \frac{i}{n+1} \leq 1 \). We then write the inequality between the \( n+1\)th and \( n\)th terms as
\[
\prod_{i=0}^{k-1} 1 - \frac{i}{n} < \prod_{i=0}^{k-1} 1 - \frac{i}{n+1} \to \prod_{i=0}^{k-1} \left( 1- \frac{i}{n} \right) \frac{x^k}{k!} < \prod_{i=0}^{k-1} \left( 1- \frac{i}{n+1} \right) \frac{x^k}{k!}
\]
The penultimate step is to show that \( \prod_{i=0}^{n} \left( 1 - \frac{i}{n+1} \right) \frac{x^{n+1}}{(n+1)!} \) is positive to prove that the \( n+1 \)th term is strictly greater than the \( n\)th term. Since \(1 \leq k \leq n \) is arbitrary, we have
\[
\sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \left( 1 - \frac{i}{n+1} \right) \right) \frac{x^k}{k!} \geq \sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \left( 1 - \frac{i}{n} \right) \right) \frac{x^k}{k!}
\]
Since \( x > 0 \), \( \frac{x^n}{(n+1)!} > 0\). Let \( 0 \leq i \leq n \). Then
\[
0 \leq i \leq n+1 \to 0 \leq \frac{i}{n+1} \leq 1 \to 0 \geq \frac{-i}{n+1} \geq -1 \to 1 \geq 1 - \frac{i}{n+1} \geq 0
\]
Therefore, every term of the product is greater than \( 0 \). Therefore, we have shown that the \( n+1 \)th term is strictly greater than the \( n \)th term. This completes the proof.
\( \square \)