Let $f(x) = x \log(1 + 1/x)$ for $0 < x < \infty$. We show that $f$ is monotonically increasing. First we rewrite $f(x) = x \log \left( \frac{x + 1}{x} \right) = x [ \log(x+1) - \log(x)]$. Taking derivatives, and doing some algebra, we get $f'(x) = [\log(x + 1) - \log(x)] - 1/(x+1)$. We wish to show that $0 < f'(x)$ everywhere. This would follow from $1/(x+1) < \log(x + 1) - \log(x)$. However, this follows from the convexity of $\log(x)$. Which says that $1/(x + \epsilon) < [\log(x + \epsilon) - \log(x)] / \epsilon$ for all $0 < x, \epsilon$.
We compute $\lim_{x \rightarrow 0} f(x)$. Taking $x_n = 1/(e^n - 1)$ we see that $f(x_n) = 1/(e^n - 1) \log(e^n) = n/(e^n - 1)$. We then have that $f(x_n) \rightarrow 0$. Since $f$ is monotonically increasing, and positive, we have that $\lim f(x) = 0$ as $x \rightarrow 0$. We compute $\lim_{x \rightarrow \infty} f(x)$. Applying concavity, we have $f(e^n) = e^n \log(1 + e^{-n}) \geq e^n \epsilon e^{-n} = \epsilon$ for all $0 < \epsilon < 1$. We then check that $f(x) = x \log(1 + 1/x) \stackrel{\star}{\leq} x (1/x) \leq 1$, where $\star$ is by applying $\log(1 + y) \leq y$. It follows that $\lim_{x \rightarrow \infty} f(x) = 1$.