We know about the functions $g_{k}(x)$ that

$$
g_{1}(x)=\frac{x^{2}}{2-x}
$$

and

$$
g_{k+1}(x)=x \frac{g_{k}(x)}{2-g_{k}(x)} \quad(k=1,2, \ldots)
$$

Determine the elements of the sequence $g_{k}^{\prime}(1)$ based on these.

Let $f(x)$ denote the function $\frac{x}{2-x}$.

$$
g_{1}(x)=x f(x)
$$

and

$$
g_{k+1}(x)=x f\left(g_{k}(x)\right), \quad(k=1,2, \ldots)
$$

Since $f(1)=1$, it follows by complete induction that $g_{1}(1)=g_{2}(1)=\ldots=g_{k}(1)=\ldots=1$.

The function $f(x)$ is differentiable at $x=1$, and thus using the previous result, we can immediately see by the method of induction that the functions $g_{k}(x)$ are also differentiable at $x=1$.

By differentiation,

$$
\begin{gathered}
g_{k+1}^{\prime}(x)=f\left(g_{k}(x)\right)+x f^{\prime}\left(g_{k}(x)\right) \quad g_{k}^{\prime}(x) \\
f^{\prime}(x)=\frac{2}{(2-x)^{2}}, \text { and thus } g_{k+1}^{\prime}(1)=1+2 g_{k}^{\prime}(1)
\end{gathered}
$$

Let us introduce the notation $h_{k}=g_{k}^{\prime}(1)+1$, then

$$
h_{k+1}=g_{k+1}^{\prime}(1)+1=2+2 g_{k}^{\prime}(1)=2\left(g_{k}(1)+1\right)=2 h_{k}
$$

From this, it follows by complete induction that

$$
h_{k+1}=2^{k} h_{1} \text {. }
$$

But $h_{1}=g_{1}^{\prime}(1)+1=4$ and thus

$g_{k}^{\prime}(1)=h_{k}-1=2^{k+1}-1, \quad(k=1,2, \ldots)$.

With this, we have determined the elements of the desired sequence.