jmc

We start by taking the first step in the Euclidean algorithm: subtract the initial two terms. Notice that $$\begin{array}{rl}{a}_{n+1}-(n+1)a_n& =(n+1)!+n+1-(n+1)(n!+n)\\ & =(n+1)!+n+1-(n+1)!-n(n+1)\\ & =-n^2+1=-(n-1)(n+1).\end{array}$$It follows that by the Euclidean Algorithm, $$\begin{array}{rl}\text{gcd}(a_n,a_{n+1})& =\text{gcd}(a_n,a_{n+1}-(n+1)a_n)\\ & =\text{gcd}(a_n,(n-1)(n+1)),\end{array}$$since the minus sign is irrelevant for calculating the gcd.

We know that $n-1$ divides $n!$, so that $n-1$ is relatively prime to $a_n=n!+n$: $$\text{gcd}(n-1,n!+n)=\text{gcd}(n-1,n)=1.$$Thus, we can ignore the factor of $n-1$ completely, and say that $$\text{gcd}(a_n,a_{n+1})=\text{gcd}(n!+n,n+1).$$Now, we have several cases, depending on whether $n+1$ is prime or composite. We also have a few edge cases to consider. The basic idea is that when $n+1$ is composite and greater than $4$, $n+1$ is a factor of $n!$, whereas when $n+1$ is prime, we can apply Wilson's Theorem.

$\text{Case 0:}$ For $n=0$, we find that $a_0=1,a_1=2$, with greatest common divisor $1$.

$\text{Case composite:}$

$\text{Subcase 1:}$ If $n+1$ is composite and can be written as the product of two distinct integers greater than $1$ (say $n+1=a\times b$, $a>b>1$), then $n+1$ divides $$n!=1\times \cdots \times b\times \cdots \times a\times \cdots \times n.$$By the same argument as before, since $n$ and $n+1$ are relatively prime, then $n!+n$ and $n+1$ are relatively prime, yielding a greatest common divisor of $1$.

$\text{Subcase 2:}$ If $n+1=p^2$ for some prime $p$, then $n!+n=(p^2-1)!+p^2-1$. If $2p<p^2-1$, then $p$ and $2p$ are both factors that appear in the expansion of $n!$, so $n+1$ divides $n!$ and the previous argument applies. For $p=2$, we can quickly check that $3!+3=9$ is relatively prime with $4$.

$\text{Case prime:}$ If $n+1=p$ for some prime $p$, then $n!+n\equiv (p-1)!+(p-1)\equiv -2(\mathrm{mod}p)$ by Wilson's Theorem. Thus, $n!+n$ is relatively prime with $n+1$ unless $n=1$, for which we obtain $a_1=2,a_2=4$, with greatest common divisor 2.

So, the largest the greatest common divisor of two consecutive terms of the sequence $a_n$ can be is $\boxed{2}$, which is achieved when we take $n=1$.