Let $0<\alpha<1$. I'm trying to find a bound on $\mathbb E\exp(\|S_n\|_1^\alpha)$ for some random walk $S_n$ on $\mathbb Z^d$. I already proved that we can bound $\mathbb P(\|S_n\|_1=k)\leq 2d\exp(-\beta k^2/n)$ for some $\beta>0$, and also the weaker result $\mathbb P(S_n=x)\leq 2d\exp(-\beta \|x\|_1^2/n)$.
I'm then asked to prove that for any $\epsilon>0$ we eventually have $$\mathbb E\exp(\|S_n\|_1^\alpha)\leq\exp(n^{\frac1{2-\alpha}+\epsilon}).$$ As a hint I was given to use that for $a_1,a_2,a_3,c>0$ and $a_2<a_3$ we have for large enough $m$ that $$m^{a_1}\exp(m^{a_2}-cm^{a_3})<\exp(2m^{a_2}-cm^{a_3})<\exp(-\frac c2m^{a_3}).$$ Expanding, this gives something like $$\mathbb E\exp(\|S_n\|_1^\alpha)=\sum_{k=0}^n\mathbb P(|\tilde S_n|=k)\exp(k^{\alpha})\leq 2\sum_{k=0}^n\exp(k^\alpha-\frac\beta nk^2).$$ Here we have simplified by noting that $\|S_n\|\stackrel{\mathcal D}=|\tilde S_n|$ where $\tilde S_n$ is a simple random walk in one dimension. Now it suffices to prove for all $\epsilon>0$, there exists an $N$ such that $n\geq N$ implies $$2\sum_{k=0}^n\exp(k^\alpha-\frac\beta nk^2)\leq\exp(n^{\frac1{2-\alpha}+\epsilon}).$$
I can't see how to proceed from here. It seems that somehow we have to fit the $\exp(k^\alpha-\frac\beta nk^2)$ term in the given exponential inequalities, but I have the feeling I'm missing some step. Any help is much appreciated.
As I indicated in the comments, since $\|S_n\|_1\leq n$, and since for any $\alpha<2$ we have $\alpha<\frac1{2-\alpha}$, it follows that $$\mathbb E\exp(\|S_n\|_1^\alpha)\leq\exp(n^\alpha)\leq\exp(n^{\frac1{2-\alpha}}).$$ Hence for all $\epsilon>0,n\geq N=0$ implies $$\mathbb E\exp(\|S_n\|_1^\alpha)\leq\exp(n^{\frac1{2-\alpha}+\epsilon}).$$ Following the approach as was hinted is not necessary.