General formula for the product $(x+1)\prod\limits_{m=1}^n (x^{2m}+1)$ Please Forgive me for any mistake in the proposal of the problem
Problem

How we can we find the general expansion of $$(x+1)\prod_{m=1}^n (x^{2m}+1),$$ where $n$ is any positive integer.

An example

For example if $n=1$ ,then the required product is $(x+1)(x^2+1)$.
  If $n=2$, then the required product is $(x+1)(x^2+1)(x^4+1)$.
   And so on the series is going like this.

My approach

I put $x^2=a$ and I decided to first solve the value after $x+1$.

Two methods
First way 

So I got a new  product which was $(a+1)(a^2+1)(a^3+1)(a^4+1)....(a^n+1)$. I tried putting the value of $n$ as $1,2,3,4$ but I could not find any pattern.

Second way

I multiplied the first term  and the last term to get $(a+1)(a^n+1)=a^{n+1} +a^n+a+1$. Then I multiplied the second last term with the second term to get$(a^2+1)(a^{n-1}+1)=a^{n+1} +a^{n-1} +a^2 +1$. 
   Then I put $a^{n+1}+1$ as $y$  and I got the general form of a term to get a new product. I again tried it for $1,2,3,4$ but I could not find any pattern.

Third way

This time I am not excluding $x+1$ and I am not taking $x^2=a$. I tried solving it as by taking $x+1=\frac{x^2-1}{x-1}$ and $x^2+1=\frac{x^4-1}{x^2-1}$ . So $x^{2n}+1=\frac{x^{4n}-1}{x^{2n}-1}$ So when we multiply them then some terms get cancelled. Example : When we solve it till $x^8+1$ , we get $\frac{(x^{12}-1)(x^{16}-1)}{(x-1)(x^6-1)}$ . I know that $x^{12}-1$ and $x^{16}-1$ can get cancelled if we try to solve it further but how will we cancel $x^6-1$?

Please help me. Please  do not use derivatives or integrals as I am not  through with them .
Forgive me for any mistake in the wording of the problem. Thanks in advance.

Apart from the initial factor, it looks (from the examples you give) as that we are speaking about
$$
P(x,n) = \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + x^{\,2\,k} } \right)} 
$$
Now,
$$
\eqalign{
  & \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + x^{\,2\,k} } \right)}  = \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + \left( {x^{\,2} } \right)^{\,k} } \right)}  = \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + y^{\,k} } \right)}  =   \cr 
  &  = \left( {1 + y} \right)\left( {1 + y^{\,2} } \right) \cdots \left( {1 + y^{\,n} } \right) = \sum\limits_{0\, \le \;m\, \le \,n\left( {n + 1} \right)/2} {f(m,n)y^{\;m} }  \cr} 
$$
so that $f(m,n)$ represents the
Number of partitions of $m$ into distinct parts ranging from $1$ to $n$
and unfortunately there is no closed form for the number of partitions of a given number (into distinct or repeated parts).
But, of course, you can manipulate the product in various ways depending on what you are aiming to. 
One way is to take the logarithm of 
$$
P(y,n) = \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + y^{\,k} } \right)} 
$$
and reach to
$$
\begin{array}{l}
 \ln P(y,n) = \ln \left( {\prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + y^{\,k} } \right)} } \right)
  = \sum\limits_{1\, \le \;k\, \le \,n} {\ln \left( {1 + y^{\,k} } \right)}  =  \\ 
  = \sum\limits_{1\, \le \;k\, \le \,n} {\ln \left( {1 + y^{\,k} } \right)}
  = \sum\limits_{1\, \le \;k\, \le \,n} {\sum\limits_{1\, \le \,j} {\left( { - 1} \right)^{\,j + 1} \frac{{y^{\,k\,j} }}{j}} }  =  \\ 
  = \sum\limits_{1\, \le \,m} {\left( {\sum\limits_{\left\{ 
  {\begin{array}{*{20}c} {k\backslash m}  \\   {1\, \le \;k\, \le \,n}  \\\end{array}} \right.} 
{\;\left( { - 1} \right)^{\,m/k + 1} k} } \right)\frac{{y^{\,m} }}{m}}  \\ 
 \end{array}
$$
Eventually, to take the derivative
$$
\frac{d}{{dy}}\ln P(y,n) = \frac{{P'(y,n)}}{{P(y,n)}} = \sum\limits_{1\, \le \,m} {\left( {\sum\limits_{\left\{ {\begin{array}{*{20}c}
   {k\backslash m}  \\
   {1\, \le \;k\, \le \,n}  \\
\end{array}} \right.} {\;\left( { - 1} \right)^{\,m/k + 1} k} } \right)y^{\,m - 1} } 
$$
which will also equal
$$
{{P'(y,n)} \over {P(y,n)}} = {{\sum\limits_{1\, \le \;j\, \le \,n} {j\,y^{\,j - 1} \prod\limits_{1\, \le \;k \ne j\, \le \,n} {\left( {1 + y^{\,k} } \right)} } } \over {\prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + y^{\,k} } \right)} }} = \sum\limits_{1\, \le \;j\, \le \,n} {{{j\,y^{\,j - 1} } \over {1 + y^{\,j} }}} 
$$
Another approach is trough the $k$-th roots of $-1$
$$
P(y,n) = \prod\limits_{1\, \le \;k\, \le \,n} {\left( {1 + y^{\,k} } \right)}  = \prod\limits_{1\, \le \;k\, \le \,n} {\prod\limits_{0\, \le \;j\, \le \,k - 1} {\left( {y - e^{\,i\,{{\left( {2j + 1} \right)} \over k}\pi } } \right)} } 
$$
etc.
