Recursion-like sequences which are hard to relate recursively Consider the sequence
\begin{align*}
   a_1&=1\\
   a_2&=2+\sqrt1\\
   a_3&=3+\sqrt{2+\sqrt1}\\
   &\kern5.5pt\vdots\\
   a_n&=n+\sqrt{n-1+\sqrt{\cdots+\sqrt{1}}}.
\end{align*}
Something like this is easy to work with inductively, since we can simply relate $a_n = n+\sqrt{a_{n-1}}$, and prove things that way. But now consider something which instead unfolds "on the inside", such as
\begin{align*}
   b_1&=1\\
   b_2&=1+\sqrt2\\
   b_3&=1+\sqrt{2+\sqrt{3}}\\
   &\kern5.5pt\vdots\\
   b_n&=1+\sqrt{2+\sqrt{\cdots+\sqrt{n}}},
\end{align*}
where it is not easy to relate $b_n$ to $b_{n-1}$. How does one work with such sequences, where we are typically interested in the same sorts of questions? Or another example, consider the sequence
\begin{align*}
   c_1&=1\\
   c_2&=1(1+2)\\
   c_3&=1(1+2(1+3))\\
   c_4&=1(1+2(1+3(1+4)))\\
   &\kern5.5pt\vdots\\
   c_n&=1(1+2(1+3(1+\cdots+(n-1)(1+n)\cdots))).
\end{align*}
Is it possible to prove inductively that $c_n = 1!+2!+\cdots+n!$, even though there is no obvious way to relate $c_n$ to $c_{n-1}$?

You can add $x$.
Let
$$\begin{align*}
   c_1'&=1x\\
   c_2'&=1(1+2x)\\
   c_3'&=1(1+2(1+3x))\\
   c_4'&=1(1+2(1+3(1+4x)))\\
   &\kern5.5pt\vdots\\
   c_n'&=1(1+2(1+3(1+\cdots+(n-1)(1+nx)\cdots))).
\end{align*}$$
So we know that $c_n'$ is $c_{n-1}'$ put $x$ as $(1+nx)$. Let $c_n'=A_n+B_nx$. So we have $A_n=A_{n-1}+B_{n-1}$ and $B_{n}=nB_{n-1}$. So you may need the recurrance relation like this.
To prove $c_n' = 1!+2!+\cdots+n!$ recursively, we can use induction to prove $A_{n}=1!+2!+\cdots+(n-1)!$ and $B_n=n!$. If this holds for $n-1$, we have $A_n=A_{n-1}+B_{n-1}=(1!+2!+\cdots+(n-2)!)+(n-1)!=1!+2!+\cdots+(n-1)!$, $B_n=nB_{n-1}=n\times (n-1)!=n!$. So $c_n$ is $c'_n$ to make $x=1$, $c_n=A_n+B_n=1!+2!+\cdots+(n-1)!+n!$.
For $b_n$, you can do it similarly (although there is no closed form). You can see $b'_n$ as functions: $b'_n(x)=1+\sqrt{2+\sqrt{\cdots+\sqrt{n+x}}}$, so we have $b'_{n+1}=b_n'(\sqrt{n+1+x})$. However, there may be no yield...
