Say you have a complete Noetherian local ring $(A,\mathfrak m_A,k_A)$ and a complete Noetherian local $A$-algebra $B$ with residue field $A$. Furthermore, suppose you can find a surjective local homomorphism $$f: k_A[[x_1,\dots,x_d]] \twoheadrightarrow B/\mathfrak m_A B$$ Then if you pick a lift $b_i \in \mathfrak m_B$ for each $i$ such that $f(x_i) = \overline{b_i}$, then you can define a local homomorphism $$g: A[[x_1,\dots,x_d]] \twoheadrightarrow B$$ taking $g: x_i \mapsto b_i$ which you can show is surjective as follows. Pick $b_0 \in B$. Then by looking mod $\mathfrak m_A$, you can find an element $c_0$ in the image of $g$ such that $b_0 - c_0 \in \mathfrak m_A B$. So you can write $$ b_0 = c_0 + \sum_i m_{1,i} b_{1,i}$$ where $m_{1,i} \in \mathfrak m_A B$ and $b_{1,i} \in B$. If you then repeat this for all of the $b_{1,i}$, and then do this forever, you then are able to express $b_0$ as a sum $\sum_n d_n$ of the form $d_n = m_1\cdots m_{k_n} c_n$, where each $c_n$ is in the image of $g$ and $m_i \in \mathfrak m_A$, such that $k_n \to \infty$ as $n \to \infty$. Since $B$ is complete, this converges, and by rearranging the terms you get the desired result. (I think this works?)
Is there a better/nicer/simpler way of proving this? I think my argument works, but it's not very satisfying, and it seems like there should be a better way of looking at it.