= Solution
Direct <integration> over <solid angle> gives
$$
1=2\pi|C|^2\int_0^\pi\sin^{2l+1}\theta\,d\theta
=4\pi|C|^2\frac{(2^l l!)^2}{(2l+1)!}.
$$
Thus $|C|=\sqrt{(2l+1)!/(4\pi)}/(2^l l!)$. Choosing the usual alternating sign convention,
$$
\boxed{Y_l^l=(-1)^l\frac{\sqrt{(2l+1)!}}{2^l l!\sqrt{4\pi}}\sin^l\theta\,e^{il\phi}.}
$$
Apply $L_-$ successively and divide by its ladder coefficient at each step. Explicitly,
$$
\boxed{Y_l^m=\sqrt{\frac{(l+m)!}{(2l)!(l-m)!}}\,\hbar^{-(l-m)}(L_-)^{l-m}Y_l^l.}
$$
The <normalization> follows because the product of the squared ladder coefficients is $\hbar^{2(l-m)}(2l)!(l-m)!/(l+m)!$. This constructs every <magnetic sublevel>, including negative $m$.
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