Please help my incorrect thinking about $(ijk)^2$ regarding quarternions

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Just a little while ago i began reading into quarternions, and was amazed by the formula $i^2 = j^2 = k^2 = ijk = -1$. After a little bit of thinking of how this could possibly be true, I found that squaring $ijk$ yields the peculiar result

$(ijk)^2 = i^2j^2k^2 = (ijk)(ijk)(ijk) = (ijk)^3$

Where is the fallacy in my line of thinking?

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$(ijk)^2\neq i^2j^2k^2$ because $i,j$, and $k$ do not commute. Indeed, $(ijk)^2=(-1)^2=1$, while $i^2j^2k^2=(-1)(-1)(-1)=-1$.