an analytic side tangent

we begin with a reminder of the binomial theorem: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] now, consider the \(n\)th term: \[ \left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^{n} \binom{n}{k} \left(1\right)^{n-k} \left(\frac{x}{n}\right)^k = \sum_{k=0}^{n} \binom{n}{k} \frac{x^k}{n^k} \] we extracte the \(k=0\) term separately: \[ \left( 1 + \frac{x}{n} \right)^n = 1 + \sum_{k=1}^{n} \frac{n!}{k!(n-k)!} \frac{x^k}{n^k} \] this part is vital, we express \(n!\) and \(n^k\) in product forms. \[ n! = \left( \prod_{i=0}^{k-1} (n-i) \right) (n-k)! \quad \text{and} \quad n^k = \prod_{i=0}^{k-1} n \] thus, we can rewrite the binomial coefficient \[ \left( 1 + \frac{x}{n} \right)^n = 1 + \sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \left(1-\frac{i}{n}\right) \right) \frac{x^k}{k!} \] for the \(n+1\)th term: \[ \left( 1 + \frac{x}{n+1} \right)^{n+1} = 1 + \sum_{k=1}^{n+1} \left( \prod_{i=0}^{k-1} \left(1-\frac{i}{n+1}\right) \right) \frac{x^k}{k!} \] expanding: \[ = 1 + \sum_{k=1}^{n} \left( \prod_{i=0}^{k-1} \left(1-\frac{i}{n+1}\right) \right) \frac{x^k}{k!} + \left( \prod_{i=0}^{n} \left(1-\frac{i}{n+1}\right) \right) \frac{x^{n+1}}{(n+1)!} \] to show strict increase, we observe: \[ n < n+1 \quad \Rightarrow \quad \frac{1}{n} > \frac{1}{n+1} \quad \Rightarrow \quad 1 - \frac{i}{n} < 1 - \frac{i}{n+1} \] thus: \[ \prod_{i=0}^{k-1} \left(1-\frac{i}{n}\right) < \prod_{i=0}^{k-1} \left(1-\frac{i}{n+1}\right) \] and for the additional term: \[ \prod_{i=0}^{n} \left(1-\frac{i}{n+1}\right) > 0 \] because each \(1 - \frac{i}{n+1}\) is positive (as \(0 \leq i \leq n\)). therefore, each term in the sum for \(n+1\) is larger, and an extra positive term is added. thus, \((1 + \frac{x}{n})^n\) is strictly increasing. \( \square \)