Asymptotics of Lambert W function

Question

I want to show $$ e^{W(n+1)}-e^{W(n)} = o(1) $$ for $n\to\infty$. Since I'm not that used to the Little-o notation I'm wondering if my reasoning is correct. I tried to prove it in the following way with the mean value theorem for $n_0\in[n,n+1]$: \begin{align} e^{W(n+1)}-e^{W(n)} &= \frac{e^{W(n+1)}-e^{W(n)}}{(n+1)-n} = \frac{d}{dn}e^{W(n)}\bigg|_{n=n_0} \\\\ &= e^{W(n_0)}\frac{W(n_0)}{n_0(1+W(n_0))} = \frac{e^{W(n_0)}}{n_0}(1+o(1)) \\\\ &= \frac{\frac{n_0}{W(n_0)}}{n_0}(1+o(1)) = \frac{1}{W(n_0)}(1+o(1)) \\\\ &= o(1)(1+o(1)) = o(1). \end{align} Is my reasoning correct or am I missing some steps?