For an LTI system, why does the Fourier transform of the impulse response give the frequency response?
Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$,
the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$
In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is 
$$\begin{align}
y(t) &amp;= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\
&amp;= \exp(j2\pi ft)\int_{-\infty}^\infty h(\tau)\exp(-j2\pi f\tau)\,\mathrm d\tau\\
&amp;= H(f)\exp(j2\pi ft),\tag{1}
\end{align}$$
where $H(f)$ is the Fourier transform of the impulse response
$h(t)$.  In words, for an LTI system with impulse response $h(t)$, the
input $\exp(j2\pi ft)$ produces output $H(f)\exp(j2\pi ft)$.
This is precisely what we have as the definition of the
frequency response of an LTI system (call
this $FR(f)$ for now): 


  for every frequency $f$, the response
  to $\exp(j2\pi ft)$ is $FR(f)\exp(j2\pi ft)$ which is just
  a (complex) constant times the input complex exponential.


But $(1)$ shows that $FR(f)$ is just $H(f)$, the Fourier
transform of the impulse response $h(t)$, which is what you
wanted to prove.