about infinitely many bounded prime gaps

Question

Let $p_n$ be the $n$'th prime $\alpha \in (0,1)$ and $e$ the euler constant. By mean value theorem one has: $$ e^{p_{n+1}^{\alpha}} = e^{c_n^{\alpha}}\frac{\alpha}{c_n^{1 - \alpha}} \cdot (p_{n+1} - p_n) + e^{p_n^{\alpha}}$$ Supposes that $\exists N \in \mathbb{n}$ such that $\forall n \geq N$ one has $p_{n+1} - p_n > D \in \mathbb{R}_+$ then it comes out that $$ e^{p_{n+1}^{\alpha}} > \alpha \cdot D \cdot \sum_{k=1}^{n} \frac{e^{c_k^{\alpha}}}{c_k^{1 - \alpha}} + C > \alpha \cdot D \cdot \sum_{k=1}^{n} \frac{e^{p_k^{\alpha}}}{p_k^{1 - \alpha}} + C$$ where $C = -\alpha\cdot D\cdot\sum_{k=1}^N\frac{e^{c_k^{\alpha}}}{c_k^{1 - \alpha}} + e^{p_N^{\alpha}}$ is a constant. But one can see that $S_\alpha(p_n) = \sum_{k=1}^{n} \frac{e^{p_k^{\alpha}}}{p_k^{1 - \alpha}} = \sum_{k=1}^{p_n} \frac{e^{p_k^{\alpha}}}{p_k^{1 - \alpha}} (\pi(k) - \pi(k-1))$ Using summation by parts it can be obtained $$ S_\alpha(p_n) = e^{(p_n+1)^{\alpha}} \frac{\pi(p_n)}{(p_n+1)^{1-\alpha}} - \int_1^{p_n} \pi(x) \left( \frac{e^{x^{\alpha}}}{x^{1-\alpha}}\right)^\prime dx \sim \frac{e^{p_n^{\alpha}} p_n^\alpha}{\log(p_n)} $$ Therefore $$ e^{p_{n+1}^{\alpha} - p_n^{\alpha}} \geq \frac{p_n^{\alpha}}{\log(p_n)} \cdot \alpha D$$ But $p_{n+1}^{\alpha} - p_n^{\alpha} \sim \frac{p_{n+1} - p_n}{p_n^{1- \alpha}}\alpha < \frac{\alpha}{p_n^{1-\alpha-\theta}}$ for $\theta > 0.525$ hence upon taking logarithm $$ \alpha \geq p_n^{1-\alpha-\theta}( \alpha\log(p_n) - \log(\log(p_n) + \log(\alpha \cdot D)) $$ which is a contradiction for $1 - \alpha - \theta > 0$ Therefore there is a sequence such that $p_{n_k+1} - p_{n_k} < D$ for $D > 0$, which is obviously not true ... where is the mistake ? Of course one can say that I cannot assume $p_{n+1} - p_n<2$ so it must be stated first $D \geq 3$ But any other mistakes?