probability question 
Suppose $X_1,X_2,...$ are i.i.d. random variables such that $E[X_j]=0$ and $P\{|X_j|>K\}=0$ for $K<\infty$. Show that $\forall t>0,\epsilon>0, P\{X_1+...+X_n\geq \epsilon n\}\leq [M(t)e^{-t\epsilon}]^n$, where $M(t) = E[e^{tX_j}]$.

Here's my attempt so far:
I note that $P\{X_1+...+X_n\geq \epsilon n\}=P\{X_1\geq\epsilon n\}+...+P\{X_n\geq\epsilon n\}$ but I'm not sure what to make of it. Maybe I should start from the right hand side and use the fact that $M(t) = E[e^{tX_j}]=E[1+tX_j+t^2X^2/2!+t^3X_j^3/3!+...]$?
 A: $P\{X_1+...+X_n\geq n\epsilon\} \le \prod_{j=1}^n P\{X_j \geq\epsilon\}$ since there are also other ways that the sum can be greater than or equal to $n\epsilon$. So now you want to show that $P\{X_j\ge\epsilon\}\le M(t)e^{-t\epsilon}$.
This is actually known as exponential Chebyshev inequality.
To derive this, note by the same reasoning
that $P\{X\ge\epsilon\}\le P\{|X|\ge\epsilon\}$
(I'm not sure whether this simplification is really necessary).
Then we need a general version of the Chebyshev inequality
thanks to @sos440 (see her/his comment below),
which is also (up to absolute value)
equivalent to its general measure-theoretic statement given on Wikipedia.
The simplification step seems to be necessary if using @sos440's derivation,
but not if one uses the general measure-theoretic statement from Wikipedia
which does not require the absolute value. 
The existence of a finite bound $K$ actually guarantees that $X$ has finite moments of each order, so that $E[e^{tX}]$ is finite.
Note, however, that Chebyshev's inequality -- and Bernstein's inequalities, for that matter -- are derived from the more basic Markov's inequality, $P\{|X|>a\}\le\frac{E[|X|]}{a}$. In particular, the latter uses $E[e^{t\sum X_j}]$ in its derivation, so your original thought of using $E[e^{tX_j}]$ might also bear fruit by applying Markov's inequality if you wanted to try your hand at a custom derivation.
