How do you determine if the series $\sum\limits_{k=1}^\infty \left(1-\frac1k\right)^{k^2}$ converges?

The root test works. Consider $$\lim \sup \sqrt[k]{\left(1 - \frac{1}{k}\right)^{k^2}} = \lim \sup \left(1 - \frac{1}{k}\right)^k = e^{-1} < 1,$$ hence the series converges.

$\begin{array}\\ (1-\frac1{k})^{k^2} &=(\frac{k-1}{k})^{k^2}\\ &=\dfrac1{(\frac{k}{k-1})^{k^2}}\\ &=\dfrac1{(1+\frac{1}{k-1})^{k^2}}\\ &=\dfrac1{((1+\frac{1}{k-1})^{k})^k}\\ &<\dfrac1{(1+\frac{k}{k-1})^k} \qquad\text{by Bernoulli}\\ &=\dfrac1{(\frac{2k-1}{k-1})^k}\\ &<\dfrac1{(\frac{2k-2}{k-1})^k}\\ &=\dfrac1{2^k}\\ \end{array} $

and the sum of this converges.