First, determine the positive integer $ n$ that yields the largest number $ n^{k}$ for some $ m$ such that $ nk=m$. Observe that $ 1^{60}<2^{30}<3^{20}>4^{15}>5^{12}$, etc. This means $ n=3$. Also, for $ k>4$, $ 2^{k}>k^{2}$, so the integer that produces the second largest $ n^{k}$ is $ 2$. Therefore, the answer is $ 3^{\lfloor\frac{1976}{3}\rfloor}\cdot2^{1976-3\lfloor\frac{1976}{3}\rfloor}=\boxed{2\cdot3^{658}}$