Several ways to prove that $\sum\limits^\infty_{n=1}\left(1-\frac1{\sqrt{n}}\right)^n$ converges

$$ \frac{\left(\,1-\frac{1}{\sqrt{n}}\,\right)^n}{e^{-\sqrt{n}}} = \frac{\left(\,\left(\,1-\frac{1}{\sqrt{n}}\,\right)^{\sqrt{n}}\,\right)^{\sqrt{n}}}{\left(\,e^{-1}\,\right)^{\sqrt{n}}} \to 1 $$ So $\sum_{n=1}^\infty \left(\,1-\frac{1}{\sqrt{n}}\,\right)^n$ and $\sum_{n=1}^{\infty} e^{-\sqrt{n}}$ converge or diverge together by the limit comparison test. Given that $$\int_1^\infty e^{-\sqrt{t}}\;dt = \left[\, -2e^{-\sqrt{x}}(\sqrt{x}+1)\,\right]_1^\infty = \frac{4}{e} < \infty$$ we conclude that both of the latter series then converge by the integral test.

Here is an elementary approach, which has the advantage of being based, first, on a basic inequality which is so useful that one should keep it in mind anyway, second, on a condensation technique which is so useful that one should keep it in mind anyway, and third, on a standard series which is so useful that one should keep it in mind anyway as well...

A basic inequality: For every $x$,

$$1-x\leqslant e^{-x}\tag{1}$$

(This stems, for example, from the fact that, the exponential function being convex, its graph is above its tangent at $x=0$.)

Now, we massage slightly the Uhr inequality $(1)$ above: if both sides are nonnegative, the inequality is preserved when raised to any positive power, hence, for every $x\leqslant1$ and every nonnegative $n$, $$(1-x)^n\leqslant e^{-nx}$$ for example, for every positive $n$, $$\left(1-\frac1{\sqrt n}\right)^n\leqslant e^{-\sqrt n}$$ hence the series of interest converges as soon as the series $$\sum_ne^{-\sqrt n}$$ converges.

A condensation technique: In words, we slice our series, the $k$th slice going from $n=k^2$ to $n=(k+1)^2-1$. Then, every term in slice $k$ is at most $e^{-k}$ and slice $k$ has $2k+1$ terms hence

$$\sum_{n=1}^\infty e^{-\sqrt n}\leqslant\sum_{k=1}^\infty (2k+1)e^{-k}\tag{2}$$

and all that remains to be shown is that the series in the RHS of $(2)$ converges.

A standard series: Perhaps the most useful series of all is the geometric series, namely the fact that, for every $|x|<1$,

$$\sum_{k=1}^\infty x^k=\frac1{1-x}\tag{3}$$

(Several simple proofs of $(3)$ exist, perhaps you already know some of them.)

Taking this for granted, note that the RHS of $(2)$ is almost a geometric series. To complete the comparison, we differentiate the geometric series term by term on $|x|<1$ (yes, this is legit), yielding $$\sum_{k=1}^\infty kx^{k-1}=\frac1{(1-x)^2}$$ We shall only keep a small part of this result, namely the fact that the series $$\sum_{k=1}^\infty x^k\qquad\text{and}\qquad\sum_{k=1}^\infty kx^k$$ both converge for every $|x|<1$.

Cauda: In particular, for $x=e^{-1}$, the series in the RHS of $(2)$ converges, qed.