Sum of i.i.d. random variables and finding an upper bound

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Problem:

Suppose that $(X_i)_{i\in\mathbb{N}^+}$ is a sequence of i.i.d. random variables. For some $n\in\mathbb{N}^+$, let $S_n=\sum_{i=1}^n X_i$. Furthermore, let $a$ be a positive constant, and $h$ be the radius of the interval containing $0$ for which the moment generating function of $X_i$ exists and has continuous derivatives.

Given that $m_X(0)=1$ and $m_X'(0)=\mathbb{E}(X)$, show that if $\mathbb{E}(X)<0$, then there exists $0<c<1$ such that $\mathbb{P}(S_n>a)\le c^n$.

Working:

For all $0<t<h$, we have that $\mathbb{P}(S_n>a)\le e^{-at}[m_X(t)]^n.$

It appears as though both the Intermediate Value Theorem and Mean Value Theorem will be useful, but I'm not entirely sure how to apply them to get the desired result.

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Since the function $m'_X$ is continuous, we can find $\delta\lt h$ such that if $|s|\lt 2\delta$, then $m'_X(s)-m'_X(0)\leqslant -\mathbb E(X)/2$ (definition of continuity with $\varepsilon:=-\mathbb E(X)/2\gt 0$). We thus have $$|s|\lt 2\delta\Rightarrow m'_X(s)\leqslant \frac{ \mathbb E(X)}2,$$ hence for $0\lt t\lt\delta$, we have $$\tag{*} m_X(t)-1=\int_0^tm'_X(s)\mathrm ds\leqslant t\mathbb E(X) /2 ,$$ from which it follows $$\mathbb P(S_n\gt a)=\mathbb P\left(e^{tS_n} \gt e^{ta}\right)\leqslant e^{-at}\left(m_X(t)\right)^n\leqslant e^{-at}\left(1+\frac t2\mathbb E(X) \right)^n.$$ Since $e^{-at}\lt 1$ and $0\lt 1+t\mathbb E(X) /2=:c\lt 1$ (by $(*)$ and the fact that $m_X(t)$ is non-negative), we are done.