Here is an elementary demonstration that uses only the definition of derivative and the simplest form of Taylor's theorem. Let's begin, therefore, by quoting these:
(1) The derivative of a function $f$ at the value $t,$ written $f^\prime(t),$ is any number for which, when $|h|$ is sufficiently small, $$\left|f(t+h)-f(t) - hf^\prime(t)\right| \le c |h|^2$$ for some number $c$ (not dependent on $h$).
Taylor's Theorem applied to the exponential function states
(2) Given $M\gt 0$, there exists a number $C_M$ for which $|e^z - 1 - z| \le C_M|z|^2$ for all complex numbers $z$ for which $|z|\le M.$
Let's generalize the question by considering functions $g$ that are integrable and bounded in size on $[-M,M]$ by, say, the number $G.$ Define
$$f_{g,X}(t) = E\left[g(X) e^{itX}\right].$$
Since the support of $X$ is on $[-M,M],$ we may compute all such expectations by integrating from $-M$ to $M.$ Therefore
$$|f_{g,X}(t)| = \left|E\left[g(X) e^{itX}\right]\right| \le E\left[\left|g(X) e^{itX}\right|\right] = E[G] = G.$$
This shows $f_{g,X}(t)$ exists and is finite for all real $t.$
Write $F$ for the distribution function of $X.$ In the following I will abuse notation by writing (for instance) "$ixg$" for the function $x\to ixg(x).$
Using only linearity of expectation, the inequality $|E[f(X)]| \le E[|f(x)|],$ and algebraic properties of the exponential, we may compute
$$\eqalign{\left|f_{g,X}(t+h) - f_{g,X}(t) - hf_{ixg,X}(t)\right|
& = \left|E\left[g(x)e^{i(t+h)x} - g(x)e^{itx} - h ixg(x) e^{itx}\right]\right| \\
&= \left|E\left[ixg(x)e^{itx}\left(e^{ihx}-1-ihx\right)\right]\right| \\
&\le E\left[\left|x\right|\left|g(x)\right|\left|e^{ihx}-1-ihx\right|\right] \\
&\le E\left[|x|GC_M\left|ihx\right|^2\right] \\
&= |h|^2 GC_M E\left[|x|^3\right] \\
& \le |h|^2 GC_M M^3.
}$$
Thus, in the definition of the derivative $(1)$ we may take $c=GC_M M^3,$ demonstrating that
$$f_{g,X}(t)^\prime = f_{ixg, X}(t).$$
By induction on the number of derivatives it is immediate that
$$f_{g,X}(t)^{(n)} = f_{(ix)^n g, X}(t).$$
Applying this to the case $g=1$ shows that $\phi(t) = f_{1,X}$ has derivatives of all orders and that $\phi^{(n)}(t) = f_{(ix)^n, X}(t),$ QED.