Direct product of two nilpotent groups is nilpotent and direct product of two solvable groups is solvable Let $G = G_{1} \times G_{2}$.
I need to prove the following two things:


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*If $G_{i}$ is nilpotent of degree $n_{i}$, $i = 1, 2$, then $G$ is nilpotent of degree $n = \max \{ n_{1}, n_{2} \}$.

*If $G_{i}$ is solvable (some people call them soluble) of degree $n_{i}$, $i = 1,2$, then $G$ is solvable of degree $n = \max \{n_{1},n_{2} \}$.


For the first one, for $G_{1}$, letting $G_{1,0}=G_{1}$, $G_{1,i+1}=G_{1,i}/C(G_{1,i})$ for $i = 0, 1, 2, \cdots$, $G_{1}$ is nilpotent if $G_{1,n_{1}}=G_{1,n_{1}-1}/C(G_{1,n_{1}-1})=\{ e \}$ for some $n_{1}$. Likewise, for $G_{2}$, letting $G_{2,0}=G_{2}$, $G_{2,i+1}=G_{2,i}/C(G_{2,i})$ for $i = 0,1,2, \cdots$, then $G_{2}$ is nilpotent if $G_{2,n_{2}}=G_{2,n_{2}-1}/C(G_{2,n_{2}-1})= \{ e\}$ for some $n_{2}$. 
So, if $G = G_{1} \times G_{2}$ were nilpotent of degree $n = \max \{n_{1}, n_{2}\}$, then $G_{\max\{ n_{1},n_{2}\}} =G_{\max\{ n_{1},n_{2}\}-1}/C(G_{\max\{ n_{1},n_{2}\}-1}) = \{ e\} $, but I can't figure out how to put the two parts together. I'm really quite clueless on how I should proceed, if this is even the best way to prove this, or even how to prove it in general. Therefore, guidance, preferably very detailed guidance, is needed.
For the second one, I was wondering if the fact that if $G = G_{1} \times G_{2}$, then $G_{1} \trianglelefteq G$, $G_{2} \trianglelefteq G$ would be needed? Although, I'm not going to lie - I'm just as clueless on this part as I was on the first part. 
Thanks! :) 

The key fact you want to use here is that $C(G_1\times G_2)=C(G_1)\times C(G_2)$ (prove this if it's not obvious to you).  And if $H_1\trianglelefteq G_1$ and $H_2\trianglelefteq G_2$, then $(G_1\times G_2)/(H_1\times H_2)\cong (G_1/H_1)\times(G_2/H_2)$.  This gives by induction on $n$ that $(G_1\times G_2)_n\cong G_{1,n}\times G_{2,n}$ for all $n$.  Since a product of groups is trivial iff each factor is trivial, this gives you the result.
For solvability the idea is very similar, but instead of looking at the operation of replacing $G$ by $G/C(G)$, you are looking at the operation of replacing $G$ by $[G,G]$.  So you want to prove that $[G_1\times G_2,G_1\times G_2]=[G_1,G_1]\times[G_2,G_2]$, and then make a similar induction argument.
