Let $L = \liminf_{m \rightarrow \infty} a_m$. This will actually be the limit. We first show $L$ is finite. To see this, multiply ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $1 \leq n < m$. You then get
$$a_m \leq e^{\sum_{n=1}^{m-1}{1 \over n^2}}a_1$$
$$< e^{\pi^2 \over 6}a_1$$
Hence the sequence is bounded and $L$ is finite.
Let $\epsilon > 0$. Since $\liminf_{m \rightarrow \infty} a_m = \sup_n \inf_{m \geq n} a_m$, for each $n, \inf_{m \geq n} a_n \leq L$ and we can choose $m$ arbitrarily large such that $a_m \leq L + \epsilon$. Suppose $M > m$.
Then multiplying ${a_{n+1} \over a_n} < 1 + {1 \over n^2} < e^{1 \over n^2}$ together for $M > n \geq m$, you get
$$a_M < e^{\sum_{n=m}^{M-1}{1 \over n^2}}a_m$$
$$< e^{\sum_{n=m}^{\infty}{1 \over n^2}}a_m$$
Given $\epsilon > 0$, if $m$ is large enough the sum $\sum_{n=m}^{\infty}{1 \over n^2}$ can be made less than $\epsilon$. Thus for any $M > m$ you have
$$a_M < (L + \epsilon) e^{\epsilon}$$
Since $\epsilon$ was arbitrary you therefore have
$\limsup_{m \rightarrow \infty}a_m \leq L$.
Since $L$ was defined as $\liminf_{m \rightarrow \infty} a_m$, the overall limit exists and equals $L$.