Question: I am given a sequence $a\_1=\ln(5)$ and $a\_n=\ln\left( \frac{e^{2a\_{n-1}}+16}{10}\right)$, for $n\geq 2$ and I am asked to show that $a\_n \geq 0$ for all $n \geq 1$.

I can do this easily with mathematical induction.

Base case: $\ln(5) \geq0$

Induction hypothesis: $a\_k \geq 0$ for all $n=k$

Induction step:

$a\_k \geq 0$

$2a\_k \geq 0$

$e^{2a\_k} \geq 1$

$e^{2a\_k} + 16 \geq 17$

$\frac{e^{2a\_k} + 16}{10} \geq \frac{17}{10}$

$\ln\left( \frac{e^{2\_{a\_k}}+16}{10}\right) \geq \ln \left( \frac{17}{10} \right) \geq 0$

$a\_{k+1} \geq 0$

So because I have shown that my induction hypothesis is true for all $n=k+1$, I'm done.

HOWEVER, this is supposedly possible to do WITHOUT mathematical induction. How would I go about doing that? For context, this is an intro level calculus course that has not covered the concept of a limit yet.

Answer: 
Yes, that all looks good to me, except that you have got $X$ and $Y$ the wrong way round. You don't need to compute the conditional probabilities to show independence (though that's certainly a valid way to do it). Instead, simply multiply any two of the marginal probabilities in the first table together, and check you don't get the corresponding entry in the table. For example $P(Y=6)=\frac{11}{36}$ and $P(X=6)=\frac{1}{36}$ (in the corrected table) but $P(X=Y=6)=\frac1{36}\neq\frac{11}{36}\times\frac{1}{36}$.

In fact it's even simpler: $P(X=4)>0$ and $P(Y=3)>0$ but $P(X=4,Y=3)=0$ so they can't be independent.
A superficial analysis, 

if $X = 6$,then $Y$ must be $6.$ While if $X = 1, Y$ could be anywhere in $[1,6]$
Hence $X,Y$ cannot be independent.