= Solution
The <weak charged current> creates a left-handed <neutrino> and a right-handed outgoing positron in the massless limit. In the parent's rest frame their momenta are opposite, so these two fixed <helicities> give total angular-momentum projection of magnitude one along their flight axis. A spin-one <W boson> can supply that projection, and its leptonic width remains nonzero:
$$
\boxed{\Gamma(W^+\to e^+\nu_e)\big|_{m_e=0}=\frac{G_FM_W^3}{6\pi\sqrt2}.}
$$
A spin-zero <pion> cannot produce the same helicity configuration. The forbidden component is rescued only by a charged-lepton mass insertion that reverses its <helicity>, giving <helicity suppression>.
This can also be seen directly from the <pion decay constant>. Lorentz symmetry gives a vacuum-to-pion <axial current> matrix element proportional to $f_\pi p_\pi^\mu$. Contracting $p_\pi=k_e+k_\nu$ with the leptonic current gives
$$
\mathcal M_\pi\ \propto\ f_\pi\bar u_\nu(\not k_\nu+\not k_e)(1-\gamma^5)v_e
=-m_e f_\pi\bar u_\nu(1+\gamma^5)v_e,
$$
using the massless neutrino equation and $(\not k_e+m_e)v_e=0$. Hence \b[$\mathcal M_\pi\propto m_e$ and $\Gamma_\pi\propto m_e^2\to0$]. The <W boson> polarization vector is not constrained to be proportional to the total momentum, so its amplitude has no analogous compulsory mass factor.
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