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<p>There is a formula to approximately calculate <img alt="n!" class="ieq" height="14" src="image/equation/n%21" width="16"/> by Stirling:</p>
<div class="eq"><img alt="N!≈√{2⋅π⋅N}⋅(÷{N}{e})^N" class="eq" height="51" src="image/equation/N%21%E2%89%88%E2%88%9A%7B2%E2%8B%85%CF%80%E2%8B%85N%7D%E2%8B%85%28%C3%B7Ne%29%5EN" width="210"/></div>
<p>In logarithms, we get:</p>
<div class="eq"><img alt="\ln N!=\ln √{2⋅π⋅N}+N⋅\ln N-N⋅\ln e" class="eq" height="21" src="image/equation/ln_N%21%3Dln_%E2%88%9A%7B2%E2%8B%85%CF%80%E2%8B%85N%7D%2BN%E2%8B%85ln_N-N%E2%8B%85ln_e" width="347"/></div>
<div class="eq"><img alt="\ln N!=\ln √{2⋅π⋅N}+N⋅\ln N-N" class="eq" height="21" src="image/equation/ln_N%21%3Dln_%E2%88%9A%7B2%E2%8B%85%CF%80%E2%8B%85N%7D%2BN%E2%8B%85ln_N-N" width="304"/></div>
<div class="eq"><img alt="\ln N!=÷{1}{2}⋅\ln 2⋅π⋅N+N⋅\ln N-N" class="eq" height="40" src="image/equation/ln_N%21%3D%C3%B712%E2%8B%85ln_2%E2%8B%85%CF%80%E2%8B%85N%2BN%E2%8B%85ln_N-N" width="318"/></div>
<div class="eq"><img alt="\ln N!=÷{1}{2}⋅(\ln 2+\ln π+\ln N)+N⋅\ln N-N" class="eq" height="40" src="image/equation/ln_N%21%3D%C3%B712%E2%8B%85%28ln_2%2Bln_%CF%80%2Bln_N%29%2BN%E2%8B%85ln_N-N" width="390"/></div>
<p>For very big N, we approximately get:</p>
<div class="eq important"><img alt="\ln N!≈N⋅\ln N-N" class="eq" height="14" src="image/equation/ln_N%21%E2%89%88N%E2%8B%85ln_N-N" width="178"/></div>
