I feel that the answer to the title quesiton is "yes". However, I tried using different bounds on such least common multiples to prove this with no luck. Any input on this is highly appreciated.

The answer is "no", the limit does not exist, because (now I'm just collecting the comments already made...) consider the series $(2p_n1)_n$, where $p_1,p_2,\dots$ denote all primes. We have $$\operatorname{lcm}(1,\dots,2p_n) = \operatorname{lcm}(1,\dots,2p_n1),$$ since $p_n$ is smaller than $2p_n$ (so $p_n$ is already in $1,\dots,2p_n$). So the quotient of the lcm is $1$. If you take the series $(p_n1)_n$, then there is a "new prime" in the numerator, so $$\operatorname{lcm}(1,\dots,p_n) = p_n \operatorname{lcm}(1,\dots,p_n1).$$ Therefore our series of lcmquotients goes to $\infty$, since there are infinitely many primes. This means the limit does not exist, the limes inferior is $1$ (since $\operatorname{lcm}(1,\dots,n)$ divides $\operatorname{lcm}(1,\dots,n+1)$), the limes superior is $\infty$. 


The limit doesn't exist, as has been pointed out. However, I'm inclined to interpret the question a bit more loosely. Let $f(n) = lcm(1, 2, \ldots n)$; then it makes sense to ask for some sort of "average" value of $f(n+1)/f(n)$. Now, $f(n+1)/f(n) = p$ if $n+1$ is a power of a prime $p$, and 1 otherwise. So the non1 values of $f(n+1)/f(n)$ get larger and larger, but also sparser and sparser, giving hope that there is some kind of average. So we first look at the mean of the first $n$ such quotients, as $n \to \infty$, $$ \lim_{n \to \infty} {1 \over n} \sum_{k=1}^n {f(k+1)/f(k)}. $$ But the sum $\sum_{k=1}^n f(k+1)/f(k)$ is at least the sum of all the primes less than $n$, which grows faster than linearly with $n$, so this limit doesn't exist. But it seems kind of silly to take the mean of quotients anyway. The more natural limit, I think, is $$ \lim_{n \to \infty} f(n)^{1/n} $$ and this in fact does exist, and has value $e$. Of course this doesn't mean that the answer to your original question is $e$. But it means that $f(n)$ is ``about'' $e^n$; more specifically it follows from the prime number theorem that $\lim_{n \to \infty} (\log f(n))/n = 1$. (I'm quoting this from the encyclopedia of integer sequences and don't remember the proof.) This gives some sense of how quickly $f(n)$ grows. 

