My understanding is that stellar fusion naturally stops at iron because it is energetically unfavourable to grow the nucleus further.
But iron is only the third most tightly-bound nucleus, nickel is number one, so shouldn't iron favourably fuse with helium?
The iron-peak elements are mostly the product of alpha capture reactions onto nuclei that begin with a similar number of neutrons and protons ($Z = N$).
The nuclear burning associated with carbon and oxygen (in type Ia supernovae) or silicon (in the cores of massive stars at the ends of their lives) is very fast or even explosive. The important reactions in determining the immediate final products are those that proceed on fast timescales.
In these cases, there is a competition between alpha capture and photodisintegration with the constraint that $Z \simeq N$, since weak, flavour-changing reactions are generally too slow to move the neutron/proton ratio far from unity before an equilibrium has been reached between alpha capture and photodisintegration.
Subject to these constraints, then $^{56}$Ni turns out to be the most stable nucleus. Further alpha captures to $^{60}$Zn (or beyond) are not favored$^*$, because the higher temperatures that would be required to overcome the greater Coulomb barrier results in photodisintegration to smaller nuclei.
Hence there is no easy route to $^{62}$Ni.
The fact that $^{56}$Fe dominates the iron-peak element abundances that are seen in the atmospheres of stars and the interstellar medium is because the $^{56}$Ni in supernovae ejecta decays to $^{56}$Co and then $^{56}$Fe on half-lives of 6 and 77 days respectively. Inside the dense core of a massive star then electron capture by $^{56}$Ni can be more rapid, but still results in $^{56}$Fe.
$*$ Note that alpha capture onto $^{56}$Ni is still energetically favourable in isolation. However, in a core made of $^{56}$Ni, then the alpha particle has to be stripped off a $^{56}$Ni nucleus first, and the net rearrangement of nucleons would be endothermic.
$^{56}Ni$ is produced in silicon-fusion stars. The fusion process doesn't "stop" at $Fe$. Several A=56 nuclides show up. See the Wiki-pedia article on :Silicon burning.
Also, Introductory Nuclear Physics by Krane, Chapter 19, Section 4.
From Wikipedia:
Because the nuclear force is stronger than the Coulomb force for atomic nuclei smaller than iron and nickel, building up these nuclei from lighter nuclei by fusion releases the extra energy from the net attraction of these particles. For larger nuclei, however, no energy is released, since the nuclear force is short-range and cannot continue to act across still larger atomic nuclei. Thus, energy is no longer released when such nuclei are made by fusion; instead, energy is absorbed in such processes.
