Let's do the math. Using Faraday's Law, integrating along a circuit, including a resistance, an ideal coil and an AC voltage source (described by ##U(t)=U_0 \exp(\mathrm{i} \omega t)##, understanding that the physical real quantities are the real part, because exponentials are much simpler to handle than cos and sin), you get
$$u+L/R \dot{u}=U_0 \exp(\mathrm{i} \omega t),$$
where ##u## is the voltage drop along the resistor.
Usually in circuit theory one considers only the stationary state and ignores the "transients" which decay exponentially on time scales ##\tau=L/R##. The stationary state is then the one, where ##u## is oscillating with the external angular frequency ##\omega##. Thus we make the ansatz
$$u(t)=u_0 \exp(\mathrm{i} \omega t),$$
where ##u_0## is a complex constant. Plugging this Ansatz into the differential equation, we get
$$(1 + \mathrm{i} \omega L/R)u_0 \exp(\mathrm{i} \omega t)=U_0 \exp(\mathrm{i} \omega t),$$
leading to
$$u_0=\frac{U_0}{1+\mathrm{i} \omega L/R} = \frac{1-\mathrm{i} \omega L/R}{1+\omega^2 L^2/R^2} U_0.$$
Now you can write
$$\frac{1-\mathrm{i} \omega L/R}{1+\omega^2 L^2/R^2}=r \exp(\mathrm{i} \varphi)$$
with
$$r=\frac{1}{\sqrt{1+\omega^2 L^2/R^2}}, \quad \varphi=-\arccos \left (\frac{1}{r} \right)<0,$$
which implies that
$$u(t)=r \exp(\mathrm{i} \omega t+\varphi)$$
or
$$\mathrm{Re} u(t)=r \cos(\omega t+\varphi).$$
Since ##\varphi<0## the phase of ##u## stays behind by this phaseshift. The current is of course
$$i(t)=\frac{1}{R} u(t)=\frac{r}{R} \cos(\omega t+\varphi).$$
Thus the phase of the current also lacks behind the source's phase by this same phaseshift.