I have some thoughts on progressing towards e being the infimum. Consider ##\limsup{a_{n+1}/a_n}##. First suppose it's larger than 1.
Since ##(1+x)/y > x/y## when x and y are positive, this means we can find ##\epsilon>0## and a subsequence for which ##((1+a_{n+1})/a_n)^n > (1+\epsilon)^n## and hence the limsup is infinity.
On the other hand if ##\limsup{a_{n+1}/a_n} <1## then the sequence decays at least as fast as a geometric sequence and must go to zero. As soon as ##a_n<1## we have ##(1+a_{n+1})/a_n > 1/a_n >1## and again the limsup must be infinity. So the only interesting thing is what happens when the limsup of this ratio is exactly 1.
Edit: now we can consider ##\limsup{(1+a_{n+1})/a_n}##. This must be at least 1 since all we've done is add 1 to the numerator. If it's larger than 1 then clearly ##\limsup{((1+a_{n+1})/a_n)^n}## is infinity. So this must also be 1. For the limsup to be 1 in both cases it must be that ##a_n\to \infty## whenever we restrict to a subsequence that makes ##a_{n_k+1}/a_{n_k}\to 1##.
So we are just interested in sequences ##a_n## going to infinity, and We can rewrite the original expression as ##(1+(a_{n+1}-a_n+1)/a_n)^n## and ,I guess the point is since we know the sequence is going to infinity, ##(a_{n+1}-a_n+1)/a_n)## is at least ##1/n##. Intuitively this is because if ##a_n## grows slower than ##n##, then we have at least ##1/a_n## and we're done. If ##a_n## grows faster, then the ##(a_{n+1}-a_n)/a_n## is at least ##1/n##. Turning this into a formal proof seems tricky and I suspect I missed a clever step to avoid a lot of this.