Euler's pentagonal theorem is the following equation: $\prod\limits_{n=1}^{+\infty}(1q^n)=\sum\limits_{m=\infty}^{+\infty}(1)^m q^{\frac{3m^2m}{2}}$ where $q<1$ is a complex number. I hope that someone will me some hints on this.

2have you looked at wikipedia? http://en.wikipedia.org/wiki/Pentagonal_number_theorem – Aug 05 '11 at 05:22

Er...You are right.I should have first looked it up in the WIKI. – user14242 Aug 05 '11 at 09:56

2See also Lecture 3 of [FuchsTabacnikov. Mathematical Omnibus](http://www.math.psu.edu/tabachni/Books/taba.pdf) – Grigory M Aug 05 '11 at 11:24
3 Answers
While there is a lot of value to the different bijective proofs known for Euler's pentagonal theorem, perhaps the proof that's easiest to see without having to draw pictures is Euler's original idea.
Define $A_N=\sum_{n=1}^{\infty}q^{N(n1)}(1q^N)\cdots (1q^{N+n1})$. Start with the identity $$\prod_{n=1}^{\infty}(1a_n)=1a_1\sum_{n=2}^{\infty}a_n(1a_1)\cdots (1a_{n1})$$ and plug in $a_n=q^n$. This gives you $\prod_{n=1}^{\infty}(1q^n)=1qq^2A_1$. Now to finish the proof of Euler's theorem you need to prove the following recursion $$A_N=1q^{2N+1}q^{3N+2}A_{N+1}.$$ This identity is equivalent to proving $$q^{2N+1}1+\sum_{n=1}^{\infty}q^{N(n1)}(1q^{N+1})\cdots(1q^{N+n1})(1q^N+q^{3N+n+1}q^{4N+2n+1})=0$$ and it starts $$(q^{2N+1}1)+(1q^N+q^{3N+2}q^{4N+3})+q^N(1q^{N+1})(1q^N+q^{3N+3}q^{4N+5})+\cdots$$ $$=q^N(1q^{N+1})(q^{2N+2}1)+q^N(1q^{N+1})(1q^N+q^{3N+3}q^{4N+5})+\cdots$$ $$=q^{2N}(1q^{N+1})(q^{2N+3}1)+\cdots$$ and if you've noticed the pattern, filling out the details shouldn't be very hard.
 1,294
 9
 12
Here is a somewhat combinatoric proof. Note that for $k>0$, the coefficient of $q^k$ in $$ \prod_{n=1}^\infty(1q^n) $$ is the number of even, increasing partitions of $k$ minus the number of odd, increasing partitions of $k$. Take $k=7$, for example. There are $3$ even, increasing partitions of $7$: $1+6$, $2+5$, and $3+4$; there are $2$ odd, increasing partitions of $7$: $7$ and $1+2+4$. Thus, the coefficient of $q^7$ is $32=1$.
In what follows, we will only be concerned with increasing partitions. There is a nice way to cancel the even partitions with the odd partitions, but it fails in a few cases. These failing cases are what give rise to the nonzero terms in the summation.
Here is the way to pair even and odd partitions. Suppose the smallest summand in a partition is $m$.
 if the $m$ largest summands are consecutive, then remove $m$ and add $1$ to the $m$ largest summands.
 if only the $j$ largest summands are consecutive $(j<m)$, then subtract $1$ from the $j$ largest summands and prepend $j$.
Note that (1) is the inverse of (2) and viceversa. Take $k=6$, for example. There are $4$ partitions, and they are paired as follows: $6\leftrightarrow 1+5$ and $1+2+3\leftrightarrow 2+4$.
Each of the two pairings above fail in exactly one way for each $m>0$,
 for the partition $m+(m+1)+...+(2m1)$ [$m$ terms totaling $(3m1)m/2$]
(this fails because $m$ is to be removed, yet it is also to be incremented)  for the partition $(m+1)+(m+2)+...+2m$ [$m$ terms totaling $(3m+1)m/2$]
(this fails because $m$ is to be prepended, yet $m+1$ is also to be decremented)
Thus, after cancellation, we have leftover partitions contributing $(1)^m q^{(3m1)m/2}$ and $(1)^m q^{(3m+1)m/2}$ for $m>0$. Noting that $(3(m)1)(m)/2=(3m+1)m/2$ and that the constant term of the product is $1=q^0$, we get $$ \prod_{n=1}^\infty(1q^n)=\sum_{m=\infty}^\infty(1)^m q^{(3m1)m/2} $$
 326,069
 34
 421
 800
There are two other proofs you might like: Andrews's proof of Jacobi triple product identity which implies Euler's Pentagonal Theorem, and has a direct bijective proof given by Sylvester (see my survey.
Another, quite different proof is due to Dyson and given here (see my survey and this popular article explaining the connection).
 1,276
 10
 19